Math in Focus

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Lesson 10.2 Finding Probability of Events detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 10 Lesson 10.2 Answer Key Finding Probability of Events

Math in Focus Grade 7 Chapter 10 Lesson 10.2 Guided Practice Answer Key

Copy and complete. Solve.

Question 1.
You roll a fair number die.

a) Find the probability of getting a four.
There are outcomes when you roll a number die. All the outcomes are equally likely.
So, the probability of getting a four is .
Answer:6 outcomes; 1/6.
Explanation:
There are 6 possible outcomes: {1, 2, 3, 4, 5, 6}
For each number, we throw a dice once then the probability of getting each number is 1/6.
The probability of number 1 is 1/6
The probability of number 2 is 1/6
The probability of number 3 is 1/6
The probability of number 4 is 1/6
The probability of number 5 is 1/6
The probability of number 6 is 1/6.

b) Find the probability of getting a seven.
It is impossible to get a seven when you roll a standard number die.
So, the probability of getting a seven is .
Answer: 0
Explanation:
Yes, it is impossible to get seven when you roll a standard number die. In a  rolling dice there will be no 7. Therefore, the probability of getting seven is 0.

Solve.

Question 2.
When you spin the spinner, what is the probability that the arrow will point to a number?

Answer: 0
Explanation:
Observe the spinner carefully,
In the spinner, all the sections are having alphabets only.
There are no numbers in the spinner. The arrow cannot show the number when we spin.

Complete.

Question 3.
Max has 4 short-sleeved shirts and 5 long-sleeved shirts in his closet. X is the event of Max randomly choosing a long-sleeved shirt. Find P(X). Express the probability as a fraction.

Event X has favorable outcomes.
P(X) = \(\frac{\text { Number of outcomes favorable to event } X}{\text { Total number of equally likely outcomes }}\). Use the formula.
=
There are long-sleeved shirts out of 9 shirts.
Answer:
The number of short-sleeved shirts=4
The number of long-sleeved shirts=5
The total number of shirts=4+5=9
Event X be choosing randomly long-sleeved shirts.
Event X=5
There are 5 long-sleeved shirts.
Here we need to find out the probability.
Formula:
The most common formula used to determine the likelihood of an event is given below:
probability=number of favourable events/total number of outcomes
P(X)=n(X)/n(S)
Where P(X)=probability
n(X)=number of favourable events
n(S)=total number of outcomes (shirts).
P(X)=5/9.
Therefore, the probability is 5/9.

Solve.

Question 4.
A box contains 28 pink ribbons and 12 green ribbons. You randomly take a ribbon from the box without looking. Find the probability of picking a pink ribbon. Express the probability as a percent.
Answer: 70%
Explanation:
The number of pink ribbons is there in a box=28
The number of green ribbons is there in a box=12
The total number of ribbons=28+12=40
The probability of finding of picking the pink ribbon=X
P(X)=n(X)/n(R)
P(X)=28/40
p(X)=0.7
Here asked that percentage. So we need to write the answer in percentage.
To find the probability of percentage:
Finally, take the answer you got and move the decimal point to the right two places or multiply the decimal by 100. Your answer will be the per cent probability that the desired outcome will take place.
Probability of percentage=0.7 x 100 = 70%
Therefore, the probability of percentage is 70%.

Question 5.
Ten cards have the following numbers printed on them: 3, 6, 9, 11, 19, 27, 35, 39, 40, and 45. A card is randomly drawn from the ten cards. Let W be the event of getting a number that is an odd number greater than 20. Let V be the event of getting a prime number.
a) List all the outcomes favourable to events W and V.
Answer:
Odd numbers: Odd numbers are the numbers that cannot be divided by 2 evenly. It cannot be divided into two separate integers evenly. If we divide an odd number by 2, then it will leave a remainder. The examples of odd numbers are 1, 3, 5, 7, etc.
Event W is the getting a number that is an odd number greater than 20
W={27, 35, 39, 45}
Prime numbers: A prime number is an integer, or whole number, greater than 1 that is only divisible by 1 and itself. In other words, a prime number only has two factors, 1 and itself.
Event V is the getting a prime number.
V= {3, 11, 9}

b) Draw a Venn diagram for the sample space and the two events. Place all possible outcomes in the Venn diagram.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
Event W is getting a number that is an odd number greater than 20
W={27, 35, 39, 45}
Event V is getting a prime number.
V= {3, 11, 9}
representation of Venn diagram:

c) Are events Wand V mutually exclusive? Explain.
Answer: Yes, W and V events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).

d) Find P(W) and P(V).
Answer:2/5; 3/10
The above-given numbers: 3, 6, 9, 11, 19, 27, 35, 39, 40, and 45.
Event W is getting a number that is an odd number greater than 20
W={27, 35, 39, 45}
P(W)=N(W)/N(T)
P(W)=4/10
P9W)=2/5
Event V is getting a prime number.
V= {3, 11, 9}
P(V)=N(V)/N(T)
P(V)=3/10
Question 6.
A letter is selected at random from the state name RHODE ISLAND. Let C be the event of getting a consonant. Let H be the event of getting a letter that comes after H in the alphabet.
a) Draw a Venn diagram for the sample space and the two events. Place all possible outcomes in the Venn diagram.
Answer:

Explanation:
Event C is getting a consonant.
Consonants are R, H, D, S, L, N
Vowels are A, E, I, O
Event H be the getting a letter that comes after H in the alphabet.
H after the alphabets are I, O
Note: A ∩ B = B ∩ A

b) Are events C and H mutually exclusive? Explain.
Answer: No, C and H are not mutually exclusive because they overlapped each other.
Two sets are said to be joint sets when they have at least one common element.

c) Find P(C) and P(H).
Answer:7/11;6/11
Explanation:
The above-given word: RHODE ISLAND
The total number of letters  in a word=11
The number of consonants= 7
P(C)=N(C)/N(T)
P(C)=7/11
Event H be the getting a letter that comes after H in the alphabet.
P(H)=N(H)/N(T)
P(H)=6/11.

Question 7.
You randomly choose a month from the twelve months in a year. Let A be the event of randomly choosing a month that has the letter a in its name.
a) Draw a Venn diagram to represent events A and A’. Give the meaning of event A’, the complement of event A.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.

A={January, February, March, April, May, August]
A’={June, July, September, October, November, December}
A’ is the complement of A.  This represents elements that are neither in set A.
A’ represents the months are having without letter ‘a’.

b) What outcomes are favourable to event A’?
Answer:
A’={June, July, September, October, November, December}
A’ is the complement of A.  This represents elements that are neither in set A.

c) Find P(A) and P(A’).
Answer:
Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen. The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
The formula for probability: The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
probability of an event to happen P(E)=number of favourable events/total number of outcomes.
A be the event of randomly choosing a month that has the letter ‘a’ in its name.
P(A)=N(A)/N(T)
P(A)=6/12
P(A)=1/2
A’ is the event having no letter in the month that is choosen.
P(A’)=N(A’)/N(T)
P(A’)=6/12
P(A’)=1/2.

Complete.

Question 8.
40% of the apples in an orchard are green and the rest of the apples are red. 5% of the red apples are rotten.
a) Copy the Venn diagram to represent the information.

Answer:
The above-given question:
Assume the total apples percentage=100%
The percentage of green apples=40%
The percentage of red apples=100-40=60%
The percentage of red apples are rotten=5%
According to the Venn diagram:
Therefore, the red apples are 60%
the green apples are 40%

b) If you pick an apple at random in the orchard, what is the probability the apple you pick is red that is not rotten? Give your answer as a decimal and as a percent.
Of all the red apples, of them are not rotten.
of 60% = • 0.60 Write percents as decimals and multiply.
= Simplify.
= % Write as a percent.
The probability of picking a red apple that is not rotten is .
Answer: 57%
Of all the red apples 95 of them are not rotten.
In simplification:
95 of 60%
=95*60/100
=57%

Solve.

Question 9.
Among the 200 jellybeans in a bag, 3 out of every 5 are blue jellybeans. The blue jellybeans consist of light blue ones and dark blue ones in the ratio 2:1.
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.

The total number of jellybeans=200
In that every 5 jellybeans we have 3 blue jellybeans. from this, we can write the number of jellybeans.
Divide 200/5=40
And multiply 3 by 40.
40*3=120.
Therefore, the number of blue jellybeans=120.
The rest of the bags means which are not blue=200-120=80.

b) What fraction of the jellybeans are light blue?
Answer:2/5
Explanation:
The ratio is already given=2:1
The light blue jellybeans ratio is 2
The dark blue jellybeans ratio is 1
We are taking 5 bags every time in that every 3 bags is blue colour only.
According to the question, the light blue was asked.
So the light blue ratio is 2 and the total is 5.
Therefore, the fraction of the light blue jellybeans is 2/5.

c) If you pick a jellybean randomly from the bag, what is the probability that a jellybean that is light blue is selected? Give your answer as a decimal.
Answer:0.4
Explanation:
The ratio is already given=2:1
The light blue jellybeans ratio is 2
The dark blue jellybeans ratio is 1
We are taking 5 bags every time in that every 3 bags is blue colour only.
According to the question, the light blue was asked.
So the light blue ratio is 2 and the total is 5.
Therefore, the fraction of the light blue jellybeans is 2/5.
P(A)=N(A)/N(T)
P(A)=2/5
P(A)=0.4
Therefore, the probability is 0.4

Math in Focus Course 2B Practice 10.2 Answer Key

Solve.

Question 1.
You roll a fair number die with faces labeled 1 to 6.
a) What is the probability of rolling an odd number?
Answer:
Probability when a dice is rolled:
total number of possible outcomes=6 (1, 2, 3, 4, 5, 6)
Probability of having 1 on the roll-up of dice:
Occurrence of 1 on top of dice/ Total number of possible outcomes of dice
= 1/6
Each of the outcomes of the dice has an equal probability of 1/6. Each outcome is equally likely to come on the top when the dice is rolled. Hence, the rolling of dice can be considered to be a fair probability scenario.
The odd numbers present on the die is 1, 3, 5
Therefore, out of the 6 total occurrences, the possible outcomes are 3.
Probability of obtaining an odd number = 3/6 = 1/2.

b) What is the probability of rolling a number less than 3?
Answer: 1/3
The probability of rolling a number less than 3 (so, 1 or 2) is 2 out of 6, or 1/3
Therefore, out of the 6 total occurrences, the possible outcomes are 2
Probability of obtaining a number less than 3=2/6=1/3.

c) What is the probability of rolling a prime number?
Answer:
total numbers that can occur=1, 2, 3, 4, 5, 6
Prime numbers that can occur in a roll=2, 3, 5
Total sample space n=6
Total favourable cases= m=3
P(prime number)=m/n=3/6
P(prime number)=1/2
therefore, the probability of rolling a prime number=1/2.

d) What is the probability of rolling a number greater than 1?
Answer:5/6
The total numbers that can occur=1, 2, 3, 4, 5, 6
The numbers that can occur in a roll that are greater than 1=2, 3, 4, 5, 6
The total sample space n =6
Total favourable cases= m = 5
P(greater than 1)=m/n=5/6
Therefore, the probability of rolling a dice greater than 1 is 5/6.

Question 2.
Abigail randomly chooses a disk from 6 green, 4 black, 2 red, and 2 white disks of the same size and shape.
a) What is the probability of getting a red disk?
Answer:
The total number of disks= n =6+4+2+2=14
The favourable cases=m=2
P(red disk)=m/n=2/14
P(red disk)=1/7
Therefore, the probability of getting a red disk=1/7.

b) What is the probability of not getting a green or white disk?
Answer:
The total number of disks= n =6+4+2+2=14
The number of disks that are green=6
The number of disks that are white=2
The disks that are not getting=m =6+2=8
P(not getting)=m/n
P(not getting)=8/14
P(not getting)=4/7
therefore, the probability of not getting green and white disks are 4/7.

c) What is the probability of getting a black disk?
Answer:
The total number of disks= n =6+4+2+2=14
The favourable cases=m=4
P(black disk)=m/n=4/14
P(black disk)=2/7
Therefore, the probability of getting a black disk=2/7.

Question 3.
A letter is randomly chosen from the word MATHEMATICS. What is the probability of choosing letter M?
Answer:
The above-given word: MATHEMATICS
The total number of letters = n = 11
The letter we need to select is ‘M’.
The favourable cases = m = 2
because we can select M two times.
P(letter M)=m/n
P(letter M)=2/11
Therefore, the probability of choosing letter M is 2/11.

Question 4.
There are 6 red marbles and 10 white marbles in a bag. What is probability of randomly choosing a white marble from the bag?
Answer:5/8
Explanation:
The number of red marbles=6
The number of white marbles=10
The total number of marbles=n=16
The favourable cases=m=10
The probability of choosing white marbles=P
P(white marbles)=m/n
P(white marbles)=10/16
P(white marbles)=5/8
Therefore, the probability of choosing white marbles are 5/8.

Question 5.
Numbers made up of two digits are formed using the digits 2, 3, and 4 with no repeating digits.
a) List all possible outcomes.
Answer:
The above-given numbers:2, 3, and 4
The possible outcomes are 23, 24, 32, 34, 42, 43

b) Find the probability of randomly forming a number greater than 32.
Answer:1/2
Explanation:
sample space={3, 4}
The total number= n= 2
According to the given numbers above, numbers obtained greater than 32 randomly=m=1 (34)
P(greater than 32)=m/n
P(greater than 32)=1/2.
Therefore, the probability of randomly forming a number greater than 32 is 1/2.

c) Find the probability of randomly forming a number divisible by 4.
Answer:1/3
The two-digit numbers formed using the given numbers 2, 3, and 4
The possible outcomes are 23, 24, 32, 34, 42, 43
Sample space={23, 24, 32, 34, 42, 43}
The total number of outcomes= n = 6
The total favourable cases= m = 2
Those cases are 24, 32 which are divisible by 4.
P(divisible by 4)=m/n
P(divisible by 4)=2/3
P(divisible by 4)=1/3.
Therefore, 1/3 is the probability of randomly forming a number divisible by 4.

Question 6.
Jack picks a letter randomly from the following list: h, i, m, o, p, q, r, t, u, and x.
The event V occurs when Jack picks a vowel.
a) Draw a Venn diagram to represent the information. Explain the meaning of the complement of event V.
Answer:

The probability of an event is a measure of the chance of occurrence of an event when an experiment is done. Complementary events occur when there are only two outcomes, for example clearing an exam or not clearing an exam. The complement means the exact opposite of an event.
Definition: For any event A, there exists another event A‘ which shows the remaining elements of the sample space S. A’ denotes complementary event of A.
: A’ = S – A.
Event A and A’ are mutually exclusive and exhaustive.
Consider the example of tossing a coin. Let P(E) denote the probability of getting a tail when a coin is tossed.
Then the probability of getting ahead is denoted by P(E’)
According to the definition, the complement of event V is the exact opposite of event V.
In complement of V (V’) all the letters are consonants, not vowels.

b) Find P(V) and P(V’).
Answer:
The number of consonants is nothing but the complement of V
The total number of letters=10
The number of vowels= V =3 (i, o, u)
P(V)=m/n
P(V)=3/10
Therefore, the probability of P(V) is 3/10
The number of consonants (V’)= 7 (h, m, p, q, r, t, x)
P(V’)=7/10
Thus the probability of the complement of V is 7/10.

Question 7.
A number is randomly selected from 1 to 20. X is the event of selecting a number divisible by 4. Y is the event of getting a prime number.
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.

Event X is selecting a number divisible by 4.
Event Y is getting a prime number.
N(X)={4, 8, 12, 16, 20}
N(Y)={2, 3, 5, 7, 11, 13, 17, 19}

b) Are events X and Y mutually exclusive? Explain.
Answer: Yes, X and Y events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).

c) Find P(X) and P(Y).
Answer:
sample space X={4, 8, 12, 16, 20}
sample space Y={2, 3, 5, 7, 11, 13, 17, 19}
The total numbers=n=20
The number of numbers divisible by 4=m=5
P(divisible by 4)=m/n
P(divisible by 4)=5/20
P(divisible by 4)=1/4
Therefore the probability of the numbers divisible by 4 is 1/4.
The number of prime numbers=8
P(prime numbers)=8/20
P(prime numbers)=2/5
Therefore, the probability of getting prime numbers is 2/5.

Question 8.
A dodecahedron number die has 12 faces. Each face is printed with one of the numbers from 1 to 12. Suppose you roll a fair dodecahedron number die and record the value on the top face. Let A be the event of rolling a number that is a multiple of 3. Let B be the event of rolling a number that is a multiple of 4.
a) Draw a Venn diagram to represent the information.
Answer:
The numbers will get when we roll a dice=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Probability of having 1 on the roll-up of dice:
Occurrence of 1 on top of dice/ Total number of possible outcomes of dice
= 1/12
Each of the outcomes of the dice has an equal probability of 1/12. Each outcome is equally likely to come on the top when the dice is rolled. Hence, the rolling of dice can be considered to be a fair probability scenario.
The possible outcome for 1 is 1/12
The possible outcome for 2 is 1/12
E(A) is the rolling number that is a multiple of 3.
E(B) in the event of rolling a number that is a multiple of 4.
The numbers multiple of 3 is 3, 6, 9, 12
The numbers multiple of 4 is 4, 8, 12

b) From the Venn diagram, tell whether events A and B are mutually exclusive. Explain your answer.
Answer: Yes, A and B events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).

c) Find P(A) and P(B).

Answer:
E(A) is the rolling number that is a multiple of 3.
E(B) in the event of rolling a number that is a multiple of 4.
The numbers multiple of 3 is 3, 6, 9, 12
The numbers multiple of 4 is 4, 8, 12
P(A)=E(A)/N(T)
P(A) is the probability of numbers that are multiples of 3
E(A)= the number of multiples that occurred in an event.
N(T) is the total number that is in the dice.
P(A)=4/12
P(A)=1/3
Therefore, the probability is 1/3.
P(B)=E(B)/N(T)
P(B) is the probability of numbers that are multiples of 4
E(B)= the number of multiples that occurred in an event.
N(T) is the total number that is in the dice.
P(B)=3/12
P(B)=1/4
Therefore, the probability is 1/4.

Question 9.
This year, some students in the Drama Club have the same first names. Name tags for the students are shown below.

One of the name tags is selected at random. Event E occurs when the name has the letter e. Event J occurs when the name tag is .
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.

E(E) is the name having with letter ‘e’.
E(J) is when the name tag occurred ‘John’.

b) List all the types of outcomes of event E’, the complement of event E.
Answer: John, Mary
Explanation:
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done. Complementary events occur when there are only two outcomes, for example clearing an exam or not clearing an exam. The complement means the exact opposite of an event.
Definition: For any event A, there exists another event A‘ which shows the remaining elements of the sample space S. A’ denotes complementary event of A.
: A’ = S – A.
Event A and A’ are mutually exclusive and exhaustive.
Consider the example of tossing a coin. Let P(E) denote the probability of getting a tail when a coin is tossed.
Then the probability of getting ahead is denoted by P(E’)
According to the definition, the complement of an event E is the exact opposite of event E.
In complement of E (E’) in a word not having letter ‘e’.

c) Are events E and J mutually exclusive? Explain your answer.
Answer: Yes, E and J events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).

d) Find P(E), P(E’), and P(J).
Answer:
sample space={Peter, Peter, Peter, James, James}
The number of event E occurs= N(E) = 5
The n\total number of names= N(T)= 9
P(E)=N(E)/N(T)
P(E)=5/9
Therefore, the probability of P(E) is 5/9.
P(E’)=N(E’)/N(T)
E’=John, Mary
The number of event E’ occurs=2
P(E’)=2/9
Therefore, the probability of E’ is 2/9.
P(J)=N(J)/N(T)
The number of event J occurs=3
P(J)=3/9
P(J)=1/3
therefore, the probability of J is 1/3.

Question 10.
Math Journal Explain why mutually exclusive events are not necessarily complementary.
Answer:
mutually exclusive events are not necessarily complementary because there will be two independent events.
Two events associated with a random experiment are said to be mutually exclusive if both cannot occur together in the same trial. Mutually-exclusive events are also known as disjoint events.
It is also important to distinguish between independent and mutually exclusive events. Independent events are those which do not depend on one another; while mutually exclusive events cannot occur together at one time.
For example: In the experiment of throwing a die, the events A = {1, 4} and B = {2, 5, 6} are mutually exclusive events.
In the same experiment, the events A = {1, 4} and C = {2, 4, 5, 6} are not mutually exclusive because, if 4 appears on the die, then it is favourable to both events A and C.
If A and B are two events, then A or B or (A ⋃ B) denotes the event of the occurrence of at least one of the events A or B.
A and B or (A ⋂ B) is the event of the occurrence of both events A and B.
If A and B happen to be mutually exclusive events, then P(A ⋂ B) = 0.

Question 11.
At a middle school, 39% of the student’s jog and 35% of the students do aerobic exercise. One out of every five students who do aerobic exercise also jogs.

a) Draw a Venn diagram to represent the information.
Answer:
Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operations performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
Jog=39%; aerobic=35%
One out of every five students who do aerobic exercise also jogs
– so if you divide students into set of 5 :
Jog=5*7=35; Aerobic=5*7=35
so 7 sets we have.
And subtract one student from every set of 7; so 39-7=32 and 35-7=28
Common students who do both are 7%

b) What per cent of the students do both activities?
Answer: 7%
Jog=39%; aerobic=35%
One out of every five students who do aerobic exercise also jogs
– so if you divide students into sets of 5 :
Jog=5*7=35; Aerobic=5*7=35
so 7 sets we have.
And subtract one student from every set of 7; so 39-7=32 and 35-7=28
Common students who do both are 7%

c) What fraction of the students only jog?
Answer:8/25
Only jog students in fractions=32/100
Therefore, the fraction of students only jog is 8/25

d) What is the probability that a randomly selected student at the middle school does neither activity? Give your answer as a decimal.
Answer:0.33
Neither do any activity=100-(32+7+28)
=100-67
=33
In decimals,
=33/100
=0.33

Question 12.
A teacher chooses a student at random from a class with 20 boys and 36 girls. 25% of the students wear glasses. 15 boys in the class do not wear glasses.
a) Draw a Venn diagram to represent the information.
Answer:
the total number of students in a class=56
The number of boys=20
The number of girls=36
The number of boys does not wear glasses=15
The remaining boys who do not wear glasses=20-15=5
Now there is a chance for 5 boys and 36 girls wearing glasses.(36+5=41)
The number of students wearing glasses=25/100*41=10.56 (nearest to 11)

b) What fraction of the students in the class are girls who do not wear glasses?
Answer:
the total number of students in a class=56
The number of boys=20
The number of girls=36
The fraction of students in the class are girls who do not wear glasses=X
25% are wearing glasses remaining 75% may not wear
X=75*36/100
X=27/56
X=1/2
c) What is the probability that a randomly selected student is a boy who wears glasses?
Answer:
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done.
P(A)=N(A)/N(T)
P(A)=5/56
P(A)=11.2
therefore, 11.2 is the probability that a randomly selected student is a boy who wears glasses

Question 13.
Alex has a pair of red socks, a pair of white socks, and a pair of black socks in his drawer. Unfortunately, the socks are not matched up with each other. Alex reaches into the drawer in the dark and pulls out two socks.
a) What ¡s the probability that the two socks are the same color?
Answer:1/5
Explanation:
Alex has pair of red socks, white socks, black socks.
He can take out the possible outcomes={RR, WW, BB}
Probability of drawing socks:
Picking the first sock of the two with the same colour has the probability of 2/6, and therefore picking the second sock with the same colour has a probability of (2−1)/(6−1)=1/5 respectively.
Occurrence of 1 pair of socks/ Total number of possible outcomes
= 1/5
Each of the outcomes of the same socks has an equal probability of 1/5. Each outcome is equally likely to come on the top when the same colour of socks is drawn. Hence, it can be considered to be a fair probability scenario.

b) What is the probability that the two socks are different colors?
Answer: 4/5
Alex has pair of red socks, white socks, black socks.
He can take out the possible outcomes of two different socks={RW, WB, RB}
Picking the first sock of the two with the different colours has the probability of 2/6, and therefore picking the second sock with the different colour has a probability of (6−2)/(6−1)=4/5 respectively.

c) Are the two events described in a) and b) complementary? Explain.
Answer: yes, the vents are complementary because two socks either match or do not match.
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done. Complementary events occur when there are only two outcomes, for example clearing an exam or not clearing an exam. The complement means the exact opposite of an event.
Definition: For any event A, there exists another event A‘ which shows the remaining elements of the sample space S. A’ denotes complementary event of A.
: A’ = S – A.
Event A and A’ are mutually exclusive and exhaustive.
Consider the example of tossing a coin. Let P(E) denote the probability of getting a tail when a coin is tossed.
Then the probability of getting ahead is denoted by P(E’)

Question 14.
A small town has a population of 3,200. 30% of the townspeople speak Italian, 20% speak French, and the rest do not speak either of these languages 360 people speak both Italian and French.
a) Draw a Venn diagram to represent the information.
Answer:

The total population of small town=3,200
The number of people who speaks italian=3200*30/100=960
The number of people who speaks French=3200*20/100=640
The number of people who speak both Italian and French=360
The rest of the people do not speak both languages=X
X=960+640+360-3200
X=1960-3200
X=1240
Therefore, 1240 people do not speak both languages.

b) What per cent of the townspeople speak only Italian?
Answer:
We already know that 30% of the people speak only Italian.
The above-given the percentage of the people who speaks only Italian.

c) If you randomly pick a person in the town and speak to the person in Italian, what is the probability that the person does not understand you?
Answer:
The total population of small town=3,200
The number of people who speaks italian=3200*30/100=960
The number of people who speaks French=3200*20/100=640
The number of people who speak both Italian and French=360
The rest of the people do not speak both languages=X
X=960+640+360-3200
X=1960-3200
X=1240
Randomly picking a person and speaking Italian to that person and asking the probability that he does not understand the Italian.
It means there is a chance of picking the person who speaks French and cannot speak both Italian and French.
So here 2 possible cases.
The total people for these 2 possible cases=640+1240=1880
P(A)=2/1880
P(A)=1/940.

Question 15.
The 6,000 oranges harvested at an orange grove are a combination of Valencia and Navel oranges. The ratio of Valencia oranges to Navel oranges is 7 : 5. The owner of the orange grove finds that 1 in every 20 Valencia oranges and 1 in every 25 Navel oranges are rotten.
a) What fraction of the oranges is not rotten?
Answer:229/240
the total number of oranges=6000
The ratio of valencia oranges=7
The ratio of Navel oranges=5
The total number of valencia oranges=X
X=6000*7/12
X=3500
1 in every 20 valencia oranges are rotten:
Valencia oranges rotten=3500/20
Valencia oranges rotten=175
Valencia oranges are not rotten=3500-175
valencia oranges not rotten=3325
Total navel oranges=6000*5/(7+5)
=6000*5/12
=2500
Navel oranges rotten=2500/25
Navel oranges rotten=100
navel oranges, not rotten=2500-100
navel oranges, not rotten=2400
Fraction of oranges not rotten:
=6000-(175+100)/6000
=6000-275/6000
=5725/6000
=229/240

b) What is the probability that a randomly selected orange is a good orange?
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
the probability that a randomly selected orange is a good orange=P(A)
oranges that are not rotten=229/240.
P(A)=229/240

c) What is the probability that a randomly selected orange is a rotten Valencia?
Answer:7/240
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Probability that a randomly selected orange is rotten valencia
valencia oranges rotten=175/2000
P(Valencia oranges rotten)=7/240

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Lesson 10.1 Defining Outcomes, Events, and Sample Space detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 10 Lesson 10.1 Answer Key Defining Outcomes, Events, and Sample Space

Hands-On Activity

Materials:

• paper
• scissors

Work in pairs.

Step 1: Make a set of four small cards with the following digits written on them.

Step 2: Pick three cards and form a 3-digit number. Record the number.
Step 3: Find all the possible 3-digit numbers that you can form. The list of numbers represents all the possible outcomes for the activity.
Step 4: Tell how many outcomes are in the sample space for this activity.

Math in Focus Grade 7 Chapter 10 Lesson 10.1 Guided Practice Answer Key

Solve.

Question 1.
Select a letter from the list of letters: A, D, E, G, K.
a) List all the possible outcomes.
Answer:
I selected the letter A. From this, we can write the possible outcomes.

The possible outcomes are: AA, AD, AE, AG, AK
If we select letter D, then the possible outcomes are:

The possible outcomes are: AD, DD, DE, DG, DK
If we take letter E, then the possible outcomes are:

The possible outcomes are: EA, Ed, EE, EG, EK
Likewise, take letter G:
The possible outcomes are: GA, GD, GE, GG, GK
Take letter K:
The possible outcomes are: KA, KD, KE, KG, KK

b) State the number of outcomes in the sample space.
Answer:
The above possible outcomes are called sample space.
A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
sample space A={AA, AD, AE, AG, AK}
sample space D={AD, DD, DE, DG, DK}
sample space E={EA, Ed, EE, EG, EK}
sample space G={GA, GD, GE, GG, GK}
sample space K={KA, KD, KE, KG, KK}
Each letter has 5 sample points. Likewise, 5 letters are there. So the total number of outcomes in the sample space are 5*5=25.
Therefore, there are 25 outcomes in the sample space.

Question 2.
Use the letter cards below to form all possible 3-letter English words.

a) List all the possible outcomes.
Answer:
The calculating of possible outcomes is a process for determining the number of possible results for an event. There are various methods for conducting this process. The best way to calculate the number of possible outcomes of an event is dependent on the type of event and the structure of the event.
By using the above-given letters a  b  c  d  d
The possible three-letter English words are:
add, bad, cab, cad, dab, dad

b) State the number of outcomes in the sample space.
Answer:
The above possible outcomes are called sample space.
A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
sample space a={add}
sample space b={bad}
sample space c={cab, cad}
sample space d={dab}
sample space d={dad}
Therefore, 6 outcomes in the sample space

Question 3.
Y is the event of choosing a prime number from a list of whole numbers from 1 to 20.
a) List the outcomes favorable to event Y.
The prime numbers up to 20 are , , , , , , ,
Y = {, , , , , , , }
Answer:
Prime numbers from 1 to 20 are the numbers that have exactly two factors and are not divisible further into a product of two natural numbers other than 1 and itself. To find whether ‘x’ is a prime number from 1 to 20, we need to check the below-mentioned conditions (all three at the same time):
Prime numbers 1 to 20 condition 1: Number should be divisible by 1 (x ÷ 1 = x)
Prime numbers 1 to 20 condition 2: Number should be divisible by itself. (x ÷ x =1)
prime numbers 1to 20 condition 3: Number must have only 2 factors 1 and the same number.
If event Y= prime numbers
You can get 2, 3, 5, 7, 11, 13, 17, 19.

b) State the number of outcomes in the sample space.
You choose from outcomes. There are outcomes in the sample space.
Answer:20; 20
The above possible outcomes are called sample space.
A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
Sample space={2, 3, 5, 7, 11, 13, 17, 19}
Therefore, there are 20 outcomes in the sample space.

Question 4.
You choose a shape from a bag containing these cardboard geometric shapes.

a) List all the favorable outcomes for the event of choosing a shape with at most 6 sides.
Answer:
Brown parallelogram, triangle, green parallelogram, square, pentagon, and hexagon.
Explanation:
Triangle: Triangle is a polygon, which is made of three sides and consists of three edges and three vertices. Also, the sum of its internal angles equals 180^o.
Square: Square is a quadrilateral where all the four sides and angles are equal and the angles at all the vertices equal to 90° each.
Parallelogram: A parallelogram is a quadrilateral with two pairs of parallel sides and opposite angles are equal in measures.
Pentagon: Pentagons can be simple or self-intersection. The properties of a simple pentagon (5-gon) are it must have five straight sides that meet to create five vertices, but do not self-intersect: Pentagons have five straight sides. Pentagons have five interior angles, which sum to 540° 540 °. The five sides do not intersect.
Hexagon: The hexagon is made up of 6 regular triangles. A closed figure which is two-dimensional and is made up of straight lines is called a polygon. A polygon is a geometric figure. In this type of geometric figures, if one has 6 sides, it is called a Hexagon.

b) List all the favorable outcomes for the event of choosing a shape with more than 4 angles.
Answer: Heptagon, octagon, pentagon, hexagon
Explanation:
Pentagon: Pentagons can be simple or self-intersection. The properties of a simple pentagon (5-gon) are it must have five straight sides that meet to create five vertices, but do not self-intersect: Pentagons have five straight sides. Pentagons have five interior angles, which sum to 540° 540 °. The five sides do not intersect.
Hexagon: The hexagon is made up of 6 regular triangles. A closed figure which is two-dimensional and is made up of straight lines is called a polygon. A polygon is a geometric figure. In this type of geometric figures, if one has 6 sides, it is called a Hexagon.
Heptagon: A heptagon has 7 sides, 7 edges, and 7 vertices. The sum of the interior angles of a heptagon is equal to 900°. The value of each interior angle of a regular heptagon is equal to 128.57°.

Octagon:
In the case of properties, we usually consider regular octagons.
– These have eight sides and eight angles.
– All the sides and all the angles are equal, respectively.
– There are a total of 20 diagonals in a regular octagon.
– The total sum of the interior angles is 1080°, where each angle is equal to 135°(135×8 = 1080)
– The Sum of all the exterior angles of the octagon is 360°, and each angle is 45°(45×8=360).

Math in Focus Course 2B Practice 10.1 Answer Key

Solve.

Question 1.
A bag contains 2 red balls and 1 green ball. A ball is taken out from the bag. What are the types of outcomes?
Answer:
The types of outcomes are red and green and we can find probability also.
Explanation:
The number of red balls=2
The number of green balls=1
The total number of balls=2+1=3
If one ball is drawn from the bag then the possible outcomes are:
P (A) = n (E)/n (S)
Where P (A) is the probability of an event “A”.
n (E) is the number of favourable outcomes.
n (S) is the total number of events in the sample space

The number of outcomes for the red ball is 1
The number of outcomes for other red ball=1
The number of outcomes for green ball=1
The total number of outcomes=1+1+1=3
P (A) = n (E)/n (S)
P (A) = 1/3.
Question 2.
A number die with faces numbered 1 to 6 is rolled once. What are the favorable outcomes for the event of getting a value that is evenly divisible by 3?
Answer:
Explanation:
– The sample space is the list of all possible outcomes.
– A dice has 6 sides which are all equally likely to be rolled.
– Each side has a different number written on it.
– we can roll a: 1, 2, 3, 4, 5, 6
– we say that the sample space for dice is {1, 2, 3, 4, 5, 6}
– The numbers that are divisible by 3 are 3 and 6.
– Therefore, the possible outcomes are {3, 6}.

Question 3.
A spinner has 5 values, as shown in the diagram. You spin the spinner and record where the spinner lands.

a) List all the outcomes in the sample space.
Answer:
The above possible outcomes are called sample space.
A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
Sample space={0, 1, 2, 3, 4}
– This spinner is split into 5 equal sections.
– One section contains 0 and another section contains 1, 2, 3, 4
– If we take 0 in one spin then the possible outcome of 0 is 1/5.
– If we take 1 in one spin then the possible outcome of 1 is 1/5.
– If we take 2 in one spin then the possible outcome of 2 is 1/5.
– If we take 3 in one spin then the possible outcome of 3 is 1/5.
– If we take 4 in one spin then the possible outcome of 4 is 1/5.

b) If event A is the event of landing on an even number, what are the outcomes favorable to event A?
Answer:
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
Sample space={0, 1, 2, 3, 4}
– This spinner is split into 5 equal sections.
In the sample space, the even numbers are 0,2, and 4.
Event A={0, 2, 4}
Total number of outcomes = n(S) = 5
A be the event of landing on even numbers
Number of favourable outcomes for each even number= n(A) = 1
that is for each even number 0, 2, 4

Question 4.
A basketball coach has 5 forwards, 2 centers, and 5 guards. E is the event of the coach picking a forward to be the team captain. How many outcomes are favorable to event E?
Answer: 5 outcomes
Explanation:
the above-given question:
The number of forwards a basketball coach has=5
The number of centres=2
The number of guards=5
The event of coach picking forwards to be the team captain=E
The number of favourable outcomes to event E=N (E).
N (E)= 5
Here asked only forwards. Therefore, 5 forwards are given so 5 possible outcomes will come.

Question 5.
A letter is selected from the letters in the name TYRANNOSAURUS REX. What is the number of possible outcomes?
Answer: 16
Explanation:

We are selecting the letters one by one.
In the word, TYRANNOSAURUS REX has 16 letters.
Therefore, the possibility of outcomes is 16.

Question 6.
A colored disk is drawn from a bag which contains the following disks.

a) If you record only the color of the disk, what are the types of outcomes?
Answer: orange, green, blue
Explanation:
If we select the disk to be drawn from a bag and the disks are having only three colours. Those three colours are orange, green, blue. Anytime you draw a disk definitely we will get those three colours.

b) If you record only the letter on the disk, what are the types of outcomes?
Answer: A, D, J, G, F, K
Explanation:
These are the possible outcomes because in the disks those letters are written.
And there is a possibility of getting 2A’s 2D’s 2K’s 1J 1G 1F.

c) If you record both the colour and the letter of the disk, what are the types of outcomes?
Answer:
If we record both colour and letter then the outcomes will be:
Orange colour with letters A, A, F, D
Green colour with letters D, K
Blue colour with letters J, G, K

Question 7.
There are 10 multiple-choice questions in a test. A student gets 1 point for every correct answer and 0 points for every wrong answer. No point is awarded for a question that the student does not try to answer.
a) What is the highest score a student could receive on the test?
Answer: 10
Explanation:
The above-given question:
The number of multiple-choice questions=10
The number of points students will get to each correct answer=10
The number of points students will get to each wrong answer=0
The highest score the students could receive on the test=X
for 10 correct answers=10 marks
for 9 correct answers=9 marks
for 8 correct answers=8 marks and so on…
compare to all the marks 10 is the highest mark.
Therefore, 10 marks will be the highest score.
b) What is the lowest possible score a student could receive?
Answer: 0
Explanation:
The number of multiple-choice questions=10
The number of points students will get to each correct answer=10
The number of points students will get to each wrong answer=0
The highest score the students could receive on the test=X
for 10 correct answers=10 marks
for 9 correct answers=9 marks
for 8 correct answers=8 marks
for 7 correct answers=7 marks
for 6 correct answers=6 marks
for 5 correct answers=5 marks
for 4 correct answers=4 marks
for 3 correct answers=3 marks
for 2 correct answers=2 marks
for 1 correct answer=1 mark
for 0 correct answers=0 marks
Compared to all the marks 0 is the lowest mark.
Therefore, 0 is the lowest mark.

c) What are the possible outcomes for a score on the multiple choice test?
Answer: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
The number of multiple-choice questions=10
The number of points students will get to each correct answer=10
The number of points students will get to each wrong answer=0
The highest score the students could receive on the test=X
for 10 correct answers=10 marks
for 9 correct answers=9 marks
for 8 correct answers=8 marks
for 7 correct answers=7 marks
for 6 correct answers=6 marks
for 5 correct answers=5 marks
for 4 correct answers=4 marks
for 3 correct answers=3 marks
for 2 correct answers=2 marks
for 1 correct answer=1 mark
for 0 correct answers=0 marks
These are all the possible outcomes to a student getting a score on the multiple-choice test.

Question 8.
A student is to be selected to play a supporting role in a drama from the list of names below.

a) X is the event that the selected student is 14 years old. List the outcomes favourable to event X.
Answer: Bob, Denise, Brenda, Benjamin
Explanation:
The above-given question:
Certain names, ages, and heights are given.
n (X) = number of possible outcomes to event X.
Once, look at the above chart and observe the age who are having 14 years. Those students are all the possible outcomes to event X.
n (X)={Bob, Denise, Brenda, Benjamin}
therefore, the number of outcomes is 4.

b) Y is the event that the selected student is at least 1.56 meters tall. List all the outcomes favorable to event Y.
Answer:

Explanation:
Once, look at the above chart and observe the height who are having at least 1.56 metres. Those students are all the possible outcomes to event Y.
n (Y)=number of possible outcomes to event Y.
n (y)={Michael, Margaret, Henry, Benjamin}
therefore, the number of outcomes is 4.

c) Z is the event that the selected student is at least 1.7 meters tall and at least 15 years old. How many outcomes are favorable to this event?
Answer:

Explanation:
Once, look at the above chart and observe the height and age who are having at least 1.7 metres and 15 years old. Those students are all the possible outcomes to event Z.
At least means we can select 1.7 and below 1.7 and coming to the age at least 15 years means we can select 15 and below 15 years old.
n (Z)=number of possible outcomes to event Z.
n (Z)={Bob, John, Margaret, Brenda, Henry, Benjamin}
therefore, the number of outcomes is 6.

d) W is the event that the name of the selected student has at most 5 letters. What outcomes are favorable to event W?
Answer:

Once, look at the above chart and observe the names column who are having at most 5 letters. Those students are all the possible outcomes to event W.
At most means, we can select 5 letters or more than 5 letters
n (W)= the number of possible outcomes to event W.
n (Z)={Denise, Michael, Josephine, Margaret, Timothy, Brenda, Henry, Chole, Benjamin}
therefore, the number of outcomes is 9.

Question 9.
From the list of digits, 0, 5, 7, 8, you form a 3-digit number without any repeating digits. You do not include any numbers that start with 0.
a) List all the outcomes of the sample space.
Answer:
Sample space= { 507, 508, 570, 578, 580, 587, 705, 708, 750, 758, 780, 785, 805, 807, 850, 857, 870, 875}
These are the outcomes of the 3-digit numbers not starting with the number 0.

b) List all the favorable outcomes for the event that a 3-digit number is an even number.
Answer:
Let’s assume that the event is A.
Event A is the even number which should be a 3-digit number.
Even number: Even numbers are those numbers that can be divided into two equal groups or pairs and are exactly divisible by 2.
Event A= {508, 570, 578, 580, 708, 750, 758, 780, 850, 870}

c) List all the favorable outcomes for the event that the 3-digit number is an odd number.
Answer:
Let’s assume that the event is B.
Event B is the odd number which should be a 3-digit number.
Odd numbers are the numbers that cannot be divided by 2 evenly. It cannot be divided into two separate integers evenly. If we divide an odd number by 2, then it will leave a remainder. The examples of odd numbers are 1, 3, 5, 7, etc.
Event B= {507, 587, 705, 785, 805, 807, 857, 875}

Question 10.
You pick a whole number between 1 and 100. If A is the event that the number you picked is divisible by 3 and 7, what are the outcomes favourable to event A?
Answer:
If event A=numbers are divisible by 3
you can get 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
n (A)=33
P (A)=33/100
If event A=numbers are divisible by 7
you can get 7 14 21 28 35 42 49 56 63 70 77 84 91 98
n (A)=14
p (A)=14/100
The Single Event Probability Calculator uses the following formulas: P (E) = n (E) / n (T) = (number of outcomes in the event) / (total number of possible outcomes) n (T) is the total number of possible outcomes. n (T) is the total number of possible outcomes.
P (E)= n (E) / n (T)
Threfore, P (A)=33/100
P (A)=14/100.

Question 11.
The 5 numeric tiles
are placed face down. You pick 4 tiles to form a 4-digit number. How many outcomes are favourable to the event of forming a number greater than 8,755?
Answer: 36
Explanation:
Let’s assume that the event is A.
Event A is the forming a 4-digit number greater than 8,755
we can pick 4 tiles such as 5, 6, 7, 8
Event A=  8756, 8757, 8758, 8765, 8766, 8767, 8768, 8775, 8776, 8777, 8778, 8785, 8786, 8787, 8788, 8885, 8886, 8887, 8888, and so on…

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Lesson 10.4 Developing Probability Models detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 10 Lesson 10.4 Answer Key Developing Probability Models

Math in Focus Grade 7 Chapter 10 Lesson 10.4 Guided Practice Answer Key

Complete. Solve.

Question 1.
There are eight letter tiles in a bag. The tiles are labeled with the letters from S to Z. June randomly selects a tile from the bag.
a) Define the sample space.
The sample space consists of 8 outcomes.
Sample space = {, , , , , , , }
Answer:
Definition: A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
Sample space={S, T, U, V, W, X, Y, Z}
It consists of 8 outcomes.

b) What is the probability of selecting a tile with a particular letter?
Every tile has an equal chance of being selected. The probability of selecting a particular tile with a particular letter is .
Answer:
The probability of selecting tiles with each letter=1/8
Probability formula:
To Calculate the probability of an event to occur we use this probability formula, recalling, the probability is the likelihood of an event to happen. This formula is going to help you to get the probability of any particular event.
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
If the probability of occurring an event is P(A) then the probability of not occurring an event is
P(A’) = 1- P(A)
Sample space={S, T, U, V, W, X, Y, Z}
n(S)=8
Favourable outcomes=1 for each letter
P(A)=1/8
The probability of selecting a tile S=1/8
The probability of selecting a tile T=1/8
The probability of selecting a tile U=1/8
The probability of selecting a tile V=1/8
The probability of selecting a tile W=1/8
The probability of selecting a tile X=1/8
The probability of selecting a tile Y=1/8
The probability of selecting a tile Z=1/8
The probability of selecting a particular tile with a particular letter is 1/8

c) Construct the probability model.
Answer:

The probability of selecting a tile S=1/8
The probability of selecting a tile T=1/8
The probability of selecting a tile U=1/8
The probability of selecting a tile V=1/8
The probability of selecting a tile W=1/8
The probability of selecting a tile X=1/8
The probability of selecting a tile Y=1/8
The probability of selecting a tile Z=1/8
The probability of selecting a particular tile with a particular letter is 1/8

d) Present the probability distribution in a bar graph.
Answer:
The probability graph displays a sample as a cumulative distribution, as different from the probability density graph or the histogram. The horizontal axis is the variable x and is usually linear or logarithmic. The vertical axis is a special probability scale derived from the inverse normal distribution function.

Solve.

Question 2.
A number is chosen at random from the list: 1, 4, 7, 12, 21, 25, 38, 40, 45, and 48.
a) Explain why a uniform probability model describes this situation.
Answer:
Definition: A model in which every outcome has equal probability.
Since the choice is random, each number has an equal chance of being chosen. So the probability of each outcome is the same.

b) What is the probability of choosing 25?
Answer:1/10
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Sample space={1, 4, 7, 12, 21, 25, 38, 40, 45, 48}
The number of sample space n(S)=10
Number of favourable outcomes=1 for each number
The probability of number 1 is 1/10
The probability of number 4 is 1/10 and so on… continue up to 48
Here the above-given question is probability of number 25
therefore, according to the above definition:
The probability of number 25 is 1/10.

c) T is the event of choosing a number that is a multiple of 3. List all the outcomes that are favorable to event T.
Answer: 12, 21, 45, 48
Explanation:
Sample space={1, 4, 7, 12, 21, 25, 38, 40, 45, 48}
Definition: A probability event can be defined as a set of outcomes of an experiment. In other words, an event in probability is the subset of the respective sample space.
N(T) is the event of choosing a number that is a multiple of 3.
The favourable outcomes of the multiples of 3 are 12, 21, 45, 48.

d) Find P(T).
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Sample space={1, 4, 7, 12, 21, 25, 38, 40, 45, 48}
The number of sample space n(S)=10
N(T) is the event of choosing a number that is a multiple of 3.
The favourable outcomes of the multiples of 3 are 12, 21, 45, 48.
The number of favourable outcomes n(E)=4
P(T)=4/10
P(T)=2/5
In decimals, we can write 0.4

Question 3.
There are 10 herbal tea bags of assorted flavors ¡n a jar: 4 peppermint, 2 raspberry, 3 camomile, and 1 blackberry. Suppose you randomly pick a tea bag from the jar and note the flavor.
a) Define the sample space.
Answer:
Definition: A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
Sample space={P, P, P, P, R, R, C, C, C, B}
The possible outcomes are 10.

b) What is the probability of picking a raspberry tea bag?
Answer:1/5
Explanation:
Sample space={P, P, P, P, R, R, C, C, C, B}
The possible outcomes=n(S)=10.
The probability of raspberry=n(E)/n(s)
The number of favourable outcomes of raspberry=n(E)=2
P(Raspberry)=2/10
P(raspberry)=1/5
In decimals, we can write 0.2
therefore, the probability of raspberry is 1/5 or 0.2

c) Construct the probability model of picking a tea bag. Then use a bar graph to represent the probability distribution.
Answer:
Sample space={P, P, P, P, R, R, C, C, C, B}
The possible outcomes=n(S)=10.
The probability of raspberry=n(E)/n(s)
The number of favourable outcomes of raspberry=n(E)=2
P(Raspberry)=2/10
P(raspberry)=1/5
In decimals, we can write 0.2
The probability of peppermint=4/10=2/5=0.4
The probability of Camomile=3/10=0.3
The probability of balckberry=1/10=0.1

Question 4.
Nine cards are made from each of the letters from the word “BEGINNING”. You select a card at random.
a) Find the probability of getting a letter N.
Answer:1/3
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Sample space={B, E, G, I, N, N, I, N, G}
The number of possible outcomes n(S)=9
The number of favourable outcomes of ‘N’=n(E)=3
the above-given was the probability of ‘N’.
P(N)=3/9
P(N)=1/3
therefore, the probability of getting N is 1/3.

b) Construct the probability model.
Answer:

Sample space={B, E, G, I, N, N, I, N, G}
The number of possible outcomes n(S)=9
The probability of getting B is 1/9
The probability of getting E is 1/9
The probability of getting G is 2/9
The probability of getting I  2/9
The probability of getting N is 3/9

c) What is the probability of selecting a card with a consonant?
Answer:2/3
Sample space={B, E, G, I, N, N, I, N, G}
The number of possible outcomes n(S)=9
The consonants of a given word are B, G, N
The number of favourable outcomes of consonants n(E)=6 (1+2+3)
The probability of getting B is 1/9
The probability of getting G is 2/9
The probability of getting N is 3/9
Now we have to find out the probability of consonants:
P(S)=6/9
P(S)=2/3
therefore, the probability of selecting a card with a consonant is 2/3.

Question 5.
The data below show the heights, in centimetres, of 25 tomato plants in a greenhouse.

a) Group the heights into 6 intervals: 80-89, 90-99, 100-109, 110-119, 120-129, and 130-139.
Answer:

Check the above data and write the intervals according to that.
Here count the number of frequencies and write into the box.
The interval of 80-90 is 4
The interval of 90-100 is 4
The interval of 100-110 is 7
The interval of 110-120 is 3
The interval of 120-130 is 5
The interval of 130-140 is 2

b) Find the relative frequencies of the 6 intervals. Construct the probability model.
Answer:
The interval of 80-90 is 4
The interval of 90-100 is 4
The interval of 100-110 is 7
The interval of 110-120 is 3
The interval of 120-130 is 5
The interval of 130-140 is 2
the total frequency is 4+4+7+3+5+2=25
We can find the relative frequency probability in the following way if we know the above two frequencies. The formula for a subgroup is;
Relative Frequency = Subgroup Count / Total Count
The relative frequency of interval 80-90=4/25
In decimal, we can write:0.16
The relative frequency of interval 90-100=4/25
In decimal, we can write:0.16
The relative frequency of interval 100-110=7/25
In decimal, we can write:0.28
The relative frequency of interval 110-129=3/25
In decimal, we can write:0.12
the relative frequency of interval 12-130=5/25=1/5
In decimal, we can write:0.2
The relative frequency of interval 130-140=2/25
In decimal, we can write:0.08
Probability model:

c) Represent the probability distribution in a histogram. State which type of probability distribution you have drawn.
Answer:
probability distribution:
Probability distribution yields the possible outcomes for any random event. It is also defined based on the underlying sample space as a set of possible outcomes of any random experiment. These settings could be a set of real numbers or a set of vectors or a set of any entities. It is a part of probability and statistics.

A non-uniform distribution is just any distribution where the probabilities are not the same.
It is a non-uniform probability distribution.

d) A plant is selected at random. What is the probability of selecting a plant whose height is greater than or equal to 100, but less than 120 centimeters tall?
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The probability of selecting a plant whose height is greater than or equal to 100, but less than 120 centimetres tall=P(A)
The number of favourable outcomes=n(E)=7+3=10
The total number of possible outcomes=n(S)=25
P(A)=10/25
P(A)=2/5
In decimals, we can write as 0.4
Therefore, the probability of selecting heights between 100 to 120 is 0.4 or 2/5.

Hands-On Activity

Materials
random number table

COMPARE AN EXPERIMENTAL PROBABILITY MODEL TO A THEORETICAL PROBABILITY MODEL

Work in pairs.

Suppose one of the digits from 0 to 9 is randomly selected. This is a theoretical uniform probability model. The probability distribution table and bar graph are given below.

Step 1: Choose any row or column of a random number table. Circle every other digit until you have circled 50 digits. Two rows of the random number table with sample digits are shown below:

Step 2: Use the digits you circled. Find the relative frequency for each digit from 0 to 9. Each digit’s relative frequency is the experimental probability of randomly choosing that digit. Record your results in a probability model like the one shown below.

Step 3: Present the probability distribution from Step 2 in a bar graph.

Step 4: Use a different row or column of the random number table and repeat step 1 so that you circle an additional 50 digits.

Step 5: You have now circled 100 digits. Find the relative frequency for each digit when the total number of circled digits is 100. The experimental probability of each randomly chosen digit (the relative frequency of the digit) is based on 100 observations. Record your results in a probability model like the one shown in Step 2.

Step 6: Present the probability distribution from Step 5 in a bar graph.

Math Journal Compare each of the experimental probability models you made with the theoretical probability model at the beginning of this activity. What effect does increasing the number of digits chosen have on the experimental probabilities? Which experimental probability model resembles the theoretical probability model more closely? Explain.
Answer:

Math in Focus Course 2B Practice 10.4 Answer Key

Solve.

Question 1.
You toss a fair coin and record whether it lands on heads or tails.
a) Define the sample space.
Answer:
Definition: A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
Sample space={heads, tails}

b) What is the probability of the coin landing on heads?
Answer:1/2
The answer to this is always going to be 50/50, or ½, or 50%. Every flip of the coin has an “ independent probability”, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Sample space={heads, tails}
n(E)=1
n(S)=2
P(A)=1/2.

c) Construct the probability model.
Answer:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Sample space={heads, tails}
n(E)=1
n(S)=2
P(heads)=1/2=0.5
P(tails)=1/2=0.5

d) Present the probability distribution in a bar graph.
Answer:

In Statistics, the probability distribution gives the possibility of each outcome of a random experiment or event. It provides the probabilities of different possible occurrences.

Question 2.
A fair icosahedron number die is a 20-faced number die which has values from 1 to 20 on its faces. You roll a fair icosahedron number die and record the number on the face the number die rests on when it lands.

a) Define the sample space.
Answer:
Definition: A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.

b) What is the probability of rolling a 14?
Answer:
The roll of a dice={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The number of possible outcomes=n(E)=1 for each
The total outcomes=n(S)=20
P(A)=1/20
the probability of rolling dice 1 = 1/20
The probability of rolling dice 2 = 1/20
Here asked the probability of rolling dice 1 is also 1/20.

c) Construct a probability model of all possible values.
Answer:

The probability of rolling dice 1 = 1/20
The probability of rolling dice 2 = 1/20
The probability of rolling dice 3 = 1/20
The probability of rolling dice 4 = 1/20
The probability of rolling dice 5 = 1/20
The probability of rolling dice 6 = 1/20
The probability of rolling dice 7 = 1/20
The probability of rolling dice 8 = 1/20
The probability of rolling dice 9 = 1/20
The probability of rolling dice 10 = 1/20
and so on…
The probability of rolling dice 20 = 1/20

d) If A is the event of rolling a prime number, find P(A).
Answer:
The prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, and 19
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
the number of possible outcomes=n(E)=8
The total outcomes=n(S)=20
P(A)=8/20
P(A)=2/5
therefore, the probability of rolling a prime number is 2/5.

e) If 8 is the event of rolling a number divisible by 4, find P(B).
Answer:
The roll of a dice={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
The event is the numbers divisible by 4. Those numbers are 4, 8, 12, 16, 20
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(B) = n(E) / n(S)
P(B) < 1
Here, P(B) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The number of outcomes=5
The total outcomes=20
P(B)=5/20
P(B)=1/4
therefore, the probability is 1/4 or 0.25

Question 3.
The spinner shown is used in a game. Before spinning the wheel, a player is given 50 points. The player then spins the wheel and adds the points indicated by the red arrow to the 50 points he or she was given. The spinner has an equal chance of landing on any one of the sections.

a) List all the possible outcomes of the game.
Answer:
The possible outcomes are {150, 500, 0, 250, 100, 50, 150, 500, 50, 100, 0, 250}
the number of possible outcomes are 12.

b) What is the probability of getting a total of 100 points?
Answer:
Sample space={150, 500, 0, 250, 100, 50, 150, 500, 50, 100, 0, 250}
The probability of getting a total of 100 points=P(A)
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The number of possible outcomes of 100 is n(E)=2
The total outcomes=12
P(A)=2/12
P(A)=1/6
Therefore, the probability is 1/6

c) Construct a probability model for all possible outcomes.
Answer:

The probability of point 0 is 2/12=1/6
The probability of point 50 is 2/12=1/6
The probability of point 100 is 2/12=1/6
The probability of point 150 is 2/12=1/6
The probability of point 250 is 2/12=1/6
The probability of point 500 is 2/12=1/6
And moreover, it is a uniform probability model.

d) Is the probability distribution uniform? State your reason.
Answer: Yes, it is a probability distribution uniform.
In probability theory and statistics, a probability distribution is a mathematical function that can be thought of as providing the probabilities of occurrence of different possible outcomes in an experiment. For instance, if the random variable X is used to denote the outcome of a coin toss (“the experiment”), then the probability distribution of X would take the value 0.5 for X = heads, and 0.5 for X = tails (assuming the coin is fair).

e) If G is the event of getting a total of 300 points, find P(G).
Answer:
if we add 250 and 50 or 150+150
P(G) is the event of getting 300 points.
A probability event can be defined as a set of outcomes of an experiment. In other words, an event in probability is the subset of the respective sample space.
sample space={150, 150, 50, 250}
Sample space={(150,150) (50, 250)}
P(G)=2/4
P(G)=1/2

Question 4.
Math Journal Tell whether you agree or disagree with the statement below. Explain your reasoning.
All events having the same number of outcomes have equal probability in a uniform probability distribution.
Answer: yes, I agree with the statement given.
In statistics, uniform distribution is a term used to describe a form of probability distribution where every possible outcome has an equal likelihood of happening. The probability is constant since each variable has equal chances of being the outcome.

Question 5.
A rectangular wooden block is painted gold on one large face and white on the other large face. The other four faces are painted green. After tossing the block many times. Sue finds that the block lands on the gold face one-third of the time. She also finds that the block has a 4% chance of landing on a green face.

a) Describe all the types of outcomes of tossing the block.
Answer: Gold, white, green
Sue finds that when the block is tossed 1/3rd of the block was a gold face and other face was white and a 4% chance of landing is on the green face.

b) Construct a probability model for all the outcomes. Write the probabilities as fractions. Is the model a uniform probability model?
Answer:

The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The probability of gold face=2/6=1/3
The probability of green face=4/6=2/3=0.04=4/100=1/25
The probability of white face=47/75
Here total 100%. In that 100%, the green face is 4% which is nothing but 1/25. and the remaining 75 will be gold and white face. The gold face is given 1/3 and the remaining is the white face.
And it is not a uniform probability model.

c) What is the probability that the block lands on the white face?
Answer:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The probability of gold face=1/3
The probability of green face=4/100=1/25
The probability of white face=47/75
Here total 100%. In that 100%, the green face is 4% which is nothing but 1/25. and the remaining 75 will be gold and white face. The gold face is given 1/3 and the remaining is the white face.

Question 6.
The data show the ages of 25 people in a group. A person is selected at random from the group.
a) Copy and complete the following frequency table.

Answer:
Frequency refers to the number of times an event or a value occurs.  A frequency table is a table that lists items and shows the number of times the items occur. We represent the frequency by the English alphabet ‘f’.

Count the numbers between the given intervals.
The frequency of the interval 10-20=6
The frequency of the interval 20-30=8
The frequency of the interval 30-40=6
the frequency of the interval=40-50=5

b) Copy and complete the following probability model.

Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The probability of the interval 10-20=6/25
The probability of the interval 20-30=8/25
The probability of the interval 30-40=5/25=1/5
The probability of the interval 40-50=6/25

c) Present the probability distribution in a histogram.
Answer:
The probability of the interval is 10-20=6/25=0.24
The probability of the interval 20-30=8/25=0.32
The probability of the interval 30-40=5/25=1/5=0.2
The probability of the interval 40-50=6/25=0.24

d) What is the probability that the selected person’s age is 30 or above?
Answer:
The probability of the interval 30-40=5/25=1/5=0.2
The probability of the interval 40-50=6/25=0.24
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
P(A)=5+6/25
P(A)=11/25
Therefore, the probability is 11/25

Question 7.
The table below shows 20 words taken from a novel. You randomly select a word from the 20 words.

a) Complete the following frequency table.

Answer:

Count the words and make a frequency table
The frequency of 3-letter words=7
The frequency of 4-letter words=5
The frequency of 5-letter words=2
The frequency of 6-letter words=6

b) Complete the following probability distribution table.

Answer:

The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The probability of 3-letter word=7/20
The probability of 4-letter word=5/20=1/4
The probability of 5-letter word=2/20=1/10
The probability of 6-letter word=6/20=3/10

c) Present the probability distribution in a bar graph.
Answer:
Probability distribution yields the possible outcomes for any random event. It is also defined based on the underlying sample space as a set of possible outcomes of any random experiment. These settings could be a set of real numbers or a set of vectors or a set of any entities. It is a part of probability and statistics.

d) What is the probability of selecting a word with at most 5 letters?
Answer:
“At most” is a probability statement, which means that the probability that a particular event will occur is less than the probability that it will not occur.
The probability of 3-letter word=7/20
The probability of 4-letter word=5/20=1/4
The probability of 5-letter word=2/20=1/10
the probability of selecting a word with at most 5 letters=P(A)
P(A)=7/20+5/20+2/20
P(a)14/20
P(A)=7/10
Therefore, the probability of selecting a word with at most 5 letters.

Question 8.
A quiz has three True-False questions. The correct answers, in order, are True-True-False (TTF). A student does not know any of the answers and decides to guess.
a) Give the sample space of all the possible outcomes for the student guessing the three answers.
Answer:
In the above-given question:
The order of the correct answers is True-True-False (TTF)
So if the student does not know any answer then there is a chance of guessing the answers.
The possibility of guessing answers is TFT, FTF, FFT, FTT, TFF
Sample space={TFT, FTF, FFT, FTT, TFF}
the possible outcomes are 5

b) Is this situation an example of a uniform probability model? Explain.
Answer: No, this is not a uniform probability model.
Explanation:
Uniform probability model: a model in which every outcome has equal probability.
Uniform probability models can be used to determine the probabilities of events.
Sample space={TFT, FTF, FFT, FTT, TFF}
The total number of outcomes=5
The possible outcomes=3
c) Construct a probability distribution table for the model.
Answer:
The probability distribution table for correct order answers:

the correct answer order is TTF
The number of possible outcomes for T is 2
The number of possible outcomes for F is 1
The total number of outcomes=3
The probability of T is 2/3
The probability of F is 1/3

d) What is the probability that the student gets all three answers correct?
Answer:1/3
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
P(A)=Probability that the student gets all three answers correct.
n(E)=The possible outcomes are 1
n(S)=Total outcomes are 3
P(A)=1/3
P9A)=0.3

e) Let X be the event that the student gets only two correct answers. List all the possible outcomes of event X.
Answer:
the correct order is TTF
P(X) is the event that the student gets only two correct answers.
The possible outcomes of event X is TFF, FTF
the number of possible outcomes are 2.
sample space={TFF, FTF}.

f) Find P(X).
Answer:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
sample space={TFF, FTF}.
The number of favourable outcomes=1 for each
The total number of possible outcomes=2
P(X)=1/2
p(X)=0.5

g) Math Journal Suppose you don’t know that the correct answers, in order, are True-True-False. Would you be able to answer e) and f)? Explain.
Answer: No we cannot answer.
Explanation:
It does not give the correct order we need to assume our own and there might be a chance of the answers are wrong.
we cannot write possible outcomes without knowing the correct answers.
The probability formula is used to compute the probability of an event to occur. To recall, the likelihood of an event happening is called probability. When a random experiment is entertained, one of the first questions that come in our mind is: What is the probability that a certain event occurs? A probability is a chance of prediction. When we assume that, let’s say, x be the chances of happening an event then at the same time (1-x) are the chances for “not happening” of an event.
Similarly, if the probability of an event occurring is “a” and an independent probability is “b”, then the probability of both the event occurring is “ab”. We can use the formula to find the chances of an event happening.
The formula for the probability of an event is:
P(A)=Number of favourable outcomes/Total number of favourable outcomes.

Brain Work

In a game, you and your friend are asked to choose a number. The number is randomly selected from a set of ten numbers from 1 to 10. Your friend chooses a card from the pack. Then you choose a card from the ones remaining in the pack. You do not know your friend’s number. You win if the difference between your number and your friend’s number is at least 3.
a) For which of your friend’s numbers do you have the greatest chance of winning?
Answer:
The set of numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
If your friend picks number 1 or number 10 then you will have the greatest chance of winning.
here at least 3 differences should be there for a chance of winning. At least means x ≥ 3.

b) For which of your friend’s numbers do you have the least chance of winning?
Answer:
If your friend picks 3, 4, 5, 6, 7, 8, 9 then you will have the least chance of winning.

c) What is the probability that you will win?
Answer:56/90
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
First, consider the greater than or equal to sign then the chances are from 3, 4, 5, 6, 7, 8, 9, 10  (8 possibilities)
Next, check out greater than a symbol then the possibilities 4, 5, 6, 7, 8, 9, 10  (7 possibilities)
The number of possibilities=8*7=n(E)=56
Already friend picks one card out of 10 cards. Then the remaining cards are 9. The total cards are 10.
Multiply 9 and 10
Total number of possibilities=10*9=90
P(A)=56/90

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Lesson 10.3 Approximating Probability and Relative Frequency detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 10 Lesson 10.3 Answer Key Approximating Probability and Relative Frequency

Hands-On Activity

Materials

• a coin

FIND RELATIVE FREQUENCIES OF FLIPPING A COIN

Work in pairs.

Step 1: Flip a coin 20 times. Record whether it lands heads or tails after each flip. The table below shows that the number of times the coin landed heads and the number of times it landed tails.

The observed frequency is the number of observations of a data value. It refers to how many times a data value appears in the chance process.

Step 2: What is the observed frequency of the coin landing on heads? What is the observed frequency of the coin landing on tails?
The relative frequency of a data value is the ratio of the observed frequency to the total number of observations in a chance process. Relative frequency can be calculated from the observed frequency.

Step 3: What is the relative frequency of the coin landing on heads? What ¡s the relative frequency of the coin landing on tails?

Step 4: What is the sum of the relative frequencies?

Math in Focus Grade 7 Chapter 10 Lesson 10.3 Guided Practice Answer Key

Complete.

Question 1.
The table shows the relative frequencies for three sizes of monitors sold during a sale at a computer store. 640 monitors were sold during the sale.

Each relative frequency represents a fraction of the total number of monitors sold. The sum of the relative frequencies is 1. 0.15 + 0.55 + 0.30 = 1
a) How many 17-inch monitors were sold during the sale?
The relative frequency 0.30 means that \(\frac{30}{100}\) of the 640 monitors sold were 17-inch monitors.
• 640 =
Multiply the relative frequency for 17-inch monitors by the total number of monitors sold.
17-inch monitors were sold during the sale.
Answer: 0.30;192
Explanation:
The number of times an event occurs is called a frequency. Relative frequency is an experimental one, but not a theoretical one. Since it is an experimental one, it is possible to obtain different relative frequencies when we repeat the experiments. To calculate the frequency we need
– Frequency count for the total population
– Frequency count for a subgroup of the population
We can find the relative frequency probability in the following way if we know the above two frequencies. The formula for a subgroup is
Relative Frequency = Subgroup Count / Total Count
The relative frequency 0.30 means that \(\frac{30}{100}\) of the 640 monitors sold were 17-inch monitors: 0.30*640=192.
Therefore, 192 17-inch monitors were sold during the sale.

b) How many fewer 14-inch monitors were sold than 15-inch monitors?
• 640 = Find the number of 14-inch monitors.
• 640 = Find the number of 15-inch monitors.
– = Subtract the number of 14-inch monitors from the number of 15-inch monitors.
There are fewer 14-inch monitors sold than 15-inch monitors.
Answer:
The relative frequency of 14-inch monitor=0.15
the relative frequency of a 15-inch monitor=0.55
Now we need to find out the number of monitors that were sold.
The number of 14-inch monitors=0.15*640=96
The number of 15-inch monitors=0.55*640=352
To know the number of fewer 14-inch monitors were sold than 15-inch monitors. We need to subtract.
: 352-96=256
therefore, 256 fewer 14-inch monitors sold than 15-inch monitors.

Copy and complete. Solve.

Question 2.
Alexis and Joe caught 40 fish over the weekend. The histogram shows the masses of the fish they caught. On the histogram, the interval 14-16 includes data for fish that have a mass of at least 14 kilograms, but less than 16 kilograms.

a) Find the relative frequency for fish that have a mass of at least 8 kilograms but less than 10 kilograms. Give your answer as a percent.
There are fish that have a mass of at least 8 kilograms but less than 10 kilograms.
= Write out of 40 fish as a fraction and simplify.
∙ 100% = Multiply by 100%
The relative frequency for fish that have a mass of at least 8 kilograms but less than 10 kilograms is %.
Answer:
According to the above-given histogram:
There are 6 fish that have a mass of at least 8 kilograms but less than 10 kilograms.
Infractions, 6/40 is equal to 3/20.
To convert into percentage multiply by100%
=3/20*100
=15%
Therefore, the relative frequency for fish that have a mass of at least 8 kilograms but less than 10 kilograms is

b) Draw a relative frequency histogram using percent.
∙ 100% =
Find the relative frequency of fish that have a mass of at least 10 kilograms but less than 12 kilograms.

∙ 100% =
Find the relative frequency of fish that have a mass of at least 12 kilograms but less than 14 kilograms.

∙ 100% = Find the relative frequency of fish that have a mass of at least 14 kilograms but less than 16 kilograms.

Answer:

The relative frequency using percentage:
1. 6/40*100=15%
2. The relative frequency of fish that have a mass of at least 10 kilograms but less than 12 kilograms.
: 17/40*100=42.5%
3. The relative frequency of fish that have a mass of at least 12 kilograms but less than 14 kilograms.
: 12/40*100=30%
4. The relative frequency of fish that have a mass of at least 14 kilograms but less than 16 kilograms.
: 5/40*100=12.5%

Question 3.
Lucas made a dartboard as shown in the diagram. He threw a dart at the dartboard 100 times. He recorded the number of times the dart landed on each color. The number of times he missed hitting the dartboard was also recorded.

a) Find the relative frequency, expressed as a decimal, for each event.
= Find the relative frequency of landing on red.
= Write the fraction as a decimal.

= Find the relative of landing on yellow.
= Write the fraction as a decimal.

= Find the relative frequency of landing on blue.
= Write the fraction as a decimal.

= Find the relative frequency of misses.
= Write the fraction as a decimal.
Answer:
Formula:
Relative frequency=subgroup count/total count
The relative frequency of landing on red=10/100
Therefore, the relative frequency is 1/10
In decimal, we can write as 0.1
The relative frequency of landing on yellow=35/100
Therefore, the relative frequency is 7/20
In decimal, we can write as 0.35
The relative frequency of landing on blue=48/100
Therefore, the relative frequency is 12/25
In decimal, we can write as 0.48
The relative frequency of misses=7/100
Therefore, the relative frequency is 7/100
In decimal, we can write as 0.07

b) Explain what the relative frequency of the dart landing in the red region means.
The event of landing in the red region has a relative frequency of , which means that the dart landed in the red region about of the time.
Answer:0.1 and 10%
The event of landing in the red region has a relative frequency of 0.1, which means that the dart landed in the red region about of the time.

c) If Lucas throws the dart again, predict in which region the dart is most likely to land.
The dart is most likely to land in the region, because it has the greatest relative frequency of .
Answer: blue, o.48
Explanation:
The relative frequency of red is 0.1
The relative frequency of yellow is 0.35
The relative frequency of blue is 0.48
By comparing all the frequencies, definitely, the dart is most likely to land in the blue region because it has the highest relative frequency of 0.48.

Hands-On Activity

Materials:

• 10 counters (5 red,
• 3 blue, and 2 green)
• a paper bag

Work in pairs.

Step 1: Place the 10 counters into the bag. Shake the bag to mix the counters. Without looking into the bag, select a counter randomly. Record its color in a tally chart, and then put the counter back in the bag. Repeat this procedure 20 times.

Step 2: Combine your group’s data from Step 1 with data from other groups. Use the class data to make a relative frequency table like the one shown below.

Step 3: Find the theoretical probabilities for the colors. Copy and complete the table below.

Step 4: Compare the experimental and the theoretical probabilities for each color. What do you observe?

Math in Focus Course 2B Practice 10.3 Answer Key

Solve.

Question 1.
A library conducted a survey on 2,000 library users about the types of books they usually borrow. The table shows the relative frequencies of the types of books borrowed.

a) How many library users borrowed biography titles?
Answer:
Relative frequency=observed frequency of data value/total number of observations
Here we know the relative frequency and observed frequency data.
Number of library users borrowed biography tiles=X
this can be write as:
0.15=2000/X
Send ‘X’ on the left side and bring relative frequency to the right side.
X=2000*0.15
X=300
therefore, 300 biography titles were borrowed.

b) What percent of the library users borrowed mystery titles?
Answer:
The relative frequency of mystery tiles=0.11
The observed frequency data=2000
The per cent of the library users borrowed mystery tiles=Y
Relative frequency=observed frequency of data value/total number of observations
Y=0.11*100
Y=11%

c) What percent of the library users borrowed romance or science fiction titles?
Answer:
the relative frequency of romance=0.28
the relative frequency of science fiction titles=0.40
The percent of the library users borrowed romance or science fiction titles=Z
Z=0.28+0.40
Z=0.68
This we can write in percentages by multiplying 100
Z=0.68*100
Z=68%
therefore, 68% of the library users borrowed romance and science fiction titles.

Question 2.
A coin is tossed 66 times and lands on heads 36 times.
a) Find the relative frequency of the coin landing on heads.
Answer:
Formula:
Relative frequency=observed frequency of data value/total number of observations
The total number of observations=66
Observed frequency=36
Relative frequency=X
According to the formula, substitute the above-given values:
X=36/66
X=6/11
In decimal, we can write as 0.54

b) Find the relative frequency of the coin landing on tails.
Answer:
Relative frequency=observed frequency of data value/total number of observations
The total number of observations=66
Observed frequency for the coin landing on head=36
Observed frequency for the coin landing on tail=66-36=30
Relative frequency of the coin landing on tails=Y
Y=30/66
Y=5/11
In decimal, we can write as 0.45

Question 3.
A number die is rolled 50 times. After each roll, the result is recorded. The table gives the observed frequency for each number on the die.

a) Copy and complete the table. Write each relative frequency as a fraction.
Answer:
The total frequency=50
The observed frequencies are 11, 4, 6, 14, 8, 7
Now we have to find relative frequency:
Relative frequency=observed frequency of data value/total number of observations
The relative frequency of value 1 = 11/50
The relative frequency of value 2 = 4/50=2/25
The relative frequency of value 3 = 6/50=3/25
The relative frequency of value 4 = 14/50=7/25
The relative frequency of value 5 = 8/50=4/25
The relative frequency of value 6 = 7/50

b) Find the relative frequency for rolling a number greater than 4. Give your answer as a fraction.
Answer:3/10
Explanation:
The dice greater than 4 is 5 and 6
The observed frequency of 5 is 8
The observed frequency of 6 is 7
Now add these two frequencies
8+7=15
Relative frequency=observed frequency of data value/total number of observations
Relative frequency=15/50
Relative frequency=3/10
Therefore, 3/10 is the relative frequency for rolling a number greater than 4.

c) Find the relative frequency of rolling an odd number. Give your answer as a fraction.
Answer:1/2
Explanation:
– We will write the probability of rolling an odd number on dice as a fraction.
– The odd numbers are 1, 3, and 5
– This is 3 of the 6 sides of the dice
– The probability of rolling an odd number on a dice is 3/6
– 3/6 is the same as the 1/2
– you can expect an odd number to be rolled half of the time.
The observed frequency of 1 is 11
The observed frequency of 3 is 6
The observed frequency of 5 is 8
Now add these three frequencies
11+6+8=25
Relative frequency=observed frequency of data value/total number of observations
Relative frequency=25/50
Relative frequency=1/2

Question 4.
The ice hockey team Blue Thunder played 25 times during the winter season. The team had 14 wins, 8 losses, and 3 ties.

a) Find the relative frequency for each of the following: wins, losses, ties. Give your answers as percents.
Answer:
The number of wins a team had=14
Relative frequency of wins=14/25
In percentage, we can write as:14/25*100=56%
The number of losses a team had=8
Relative frequency of losses=8/25
In percentage, we can write as:8/25*100=32%
The number of ties a team had=3
Relative frequency of losses=3/25
In percentage, we can write as:3/25*100=12%

b) Draw a relative frequency bar graph that uses percents.
Answer:

c) What percent of the total number of games ended in a loss or a tie?
Answer:
The number of losses a team had=8
Relative frequency of losses=8/25
In percentage, we can write as:8/25*100=32%
The number of ties a team had=3
Relative frequency of losses=3/25
In percentage, we can write as:3/25*100=12%

Question 5.
You are given a deck of 52 cards showing scenes of the four seasons of the year. There are 13 cards for each season. You randomly select a card from the deck 100 times. After each selection, you record the season shown on the card and replace it in the deck for the next selection. The results are given in the table below.

a) Find the relative frequency of the appearance of each season.
Answer:
The total number of observations=100
The observed frequencies=23, 34, 19, 24
Relative frequency=observed frequency of data value/total number of observations
The relative frequency of the spring=23/100
The relative frequency of the summer=34/100=17/50
The relative frequency of the fall=19/100
The relative frequency of the winter=24/100=6/25

b) What is the experimental probability of selecting a card with a summer scene?
Answer:
Experimental probability: Experimental probability, also known as Empirical probability, is based on actual experiments and adequate recordings of the happening of events. To determine the occurrence of any event, a series of actual experiments are conducted. Experiments that do not have a fixed result are known as random experiments. The outcome of such experiments is uncertain. Random experiments are repeated multiple times to determine their likelihood. An experiment is repeated a fixed number of times and each repetition is known as a trial. Mathematically, the formula for the experimental probability is defined by;
Probability of an Event P(E) = Number of times an event occurs / Total number of trials.
The observed frequency of the summer=34
According to the above definition:
P(E)=34/100
p(E)=17/50

c) What is the theoretical probability of selecting a card with a summer scene?
Answer:
In probability, the theoretical probability is used to find the probability of an event. Theoretical probability does not require any experiments to conduct. Instead of that, we should know about the situation to find the probability of an event occurring. Mathematically, the theoretical probability is described as the number of favourable outcomes divided by the number of possible outcomes.
Probability of Event P(E) = No. of. Favourable outcomes/ No. of. Possible outcomes.
N0.of favourable outcomes of summer=1
No.of possible outcomes=4 (4 seasons)
P(E)=1/4

d) Math journal The experimental and the theoretical probabilities J of selecting a card with a summer scene are not equal. Describe some
factors that cause the two probabilities to be different.
Answer:
Some possible factors are as follows: The cards may not be well-shuffled; different people choose a card in different ways. How the cards are placed may make a difference; The theoretical probability may have assumptions that cannot be achieved from experiments.

e) Suppose that the spring and summer cards have a red background and the fall and winter cards have a black background. What is the experimental probability of selecting a card with a black background?
Answer:43/100
The total number of observations=100
The observed frequencies=23, 34, 19, 24
Relative frequency=observed frequency of data value/total number of observations
The relative frequency of the spring=23/100
The relative frequency of the summer=34/100=17/50
The relative frequency of the fall=19/100
The relative frequency of the winter=24/100=6/25
Experimental probability:
Probability of an Event P(E) = Number of times an event occurs / Total number of trials.
Now add the fall and winter cards:
Calculation:

100 is the least common multiple of denominators 100 and 25. Use it to convert to equivalent fractions with this common denominator.
Therefore, the experimental probability of selecting a card with a black background is 43/100.

f) What is the theoretical probability of selecting a card with a red background?
Answer:1/2
In the above question given that the spring and summer cards have a red background. According to that we need to calculate the probability.
In probability, the theoretical probability is used to find the probability of an event. Theoretical probability does not require any experiments to conduct. Instead of that, we should know about the situation to find the probability of an event occurring. Mathematically, the theoretical probability is described as the number of favourable outcomes divided by the number of possible outcomes.
Probability of Event P(E) = No. of. Favourable outcomes/ No. of. Possible outcomes.
The number of favourable outcomes=1
the number of possible outcomes=2
P(E)=1/2

Question 6.
A group of researchers catch and measure the length offish before releasing them back to a river. The lengths of the 50 fish are categorized in the table below.

a) Find the relative frequency of each category of fish.
Answer:
The total number of observations=50
The observed frequencies=16, 23, 11
Relative frequency=observed frequency of data value/total number of observations
The relative frequency of length of the fish (L<7)=16/50=8/25
The relative frequency of length of the fish 7less than or equal to L < 14=23/50
The relative frequency of length of the fish 14 less than or equal to L < 21=11/50

b) If the researchers catch and measure one more fish, what is the probability of catching a fish that is less than 7 inches long?
Answer:
The probability of catching fish is less than 7 and the possible outcomes are 1, 2, 3, 4, 5, 6
P(E)=N(P)/N(T)
P(E)=5/50
P(E)=1/10
Therefore, the probability of catching a fish that is less than 7 inches long is 1/10.

c) What is the probability that the next fish caught will be at least 7 inches long?
Answer:
The probability of catching fish is less than 7 and the possible outcomes are 1, 2, 3, 4, 5, 6,7
P(E)=N(P)/N(T)
P(E)=7/50
Therefore, the probability of catching a fish that is less than 7 inches long is 7/10.

d) Draw a relative frequency histogram using percent.
Answer:
The total number of observations=50
The observed frequencies=16, 23, 11
Relative frequency=observed frequency of data value/total number of observations
The relative frequency of length of the fish (L<7)=16/50=8/25
In percentages, we can write 8/25*100=8%
The relative frequency of length of the fish 7less than or equal to L < 14=23/50
In percentages, we can write 23/50*100=11.5%
The relative frequency of length of the fish 14 less than or equal to L < 21=11/50
In percentages, we can write 11/50*100=5.5%

Question 7.
A car dealership has sold 75 new cars this year. The histogram below shows frequencies for cars sold in different price ranges.

a) Find the relative frequency of the cars sold in the price range of $26,000 to $28,000. Give your answer as a percent.
Answer:
The total number of observations=75
The observed frequencies=24, 15, 21, 12, 3
Relative frequency=observed frequency of data value/total number of observations
Relative frequency=12/75=0.16
In percentages, we can write 0.16*100=16%
Therefore, the relative frequency of the cars sold in the price range of $26,000 to $28,000 is 16%.

b) Find the relative frequency of the cars sold in the price range of $20,000 to $24,000. Give your answer as a percent.
Answer:
The total number of observations=75
The observed frequencies=24, 15, 21, 12, 3
Relative frequency=observed frequency of data value/total number of observations
The relative frequency of the cars sold in the price range of $20,000 to $24,000=X
X=24+15/75
X=39/75
X=0.52
In percentages, we can write 0.52*100=52%

c) What is the probability that the next car will be sold in the price range of $28,000 to $30,000?
Answer:
the probability of the next car will be sold in the price range of $28,000 to $30,000=3/75=0.04
In percentages, we can write 0.04*100=4%.

Question 8.
A light bulb manufacturer estimates that 10% of a shipment of 600 bulbs will have a lifespan greater than or equal to 1,000 hours, but less than 2,000 hours. The manufacturer also estimates that 240 of the bulbs will have a lifespan greater than or equal to 2,000 hours, but less than 3,000 hours, and that the remaining light bulbs will have a lifespan greater than or equal to 3,000 hours, but less than 4,000 hours.
a) Copy and complete the table below.

Answer:

Explanation:
The number of bulbs will have a lifespan greater than or equal to 1,000 hours, but less than 2,000 hours
:10% of 600=60 bulbs
10*600/100=60 bulbs.
So I had written 60 in the first box.
Already given 240 bulbs are given for second box.
The remaining boxes =X
X=240+60=300
Now subtract the total bulbs and the life span of the present boxes.
X=600-300
X=300
The remaining bulbs are 300.

b) Draw a frequency histogram for the three lifespans shown in the table.
Answer:

A histogram is a graphical representation of a grouped frequency distribution with continuous classes. It is an area diagram and can be defined as a set of rectangles with bases along with the intervals between class boundaries and with areas proportional to frequencies in the corresponding classes. In such representations, all the rectangles are adjacent since the base covers the intervals between class boundaries. The heights of rectangles are proportional to corresponding frequencies of similar classes and for different classes, the heights will be proportional to corresponding frequency densities.
In other words, a histogram is a diagram involving rectangles whose area is proportional to the frequency of a variable and width is equal to the class interval.

c) Find the relative frequency for the bulbs with each of the lifespans. Then draw a relative frequency histogram using percents.
Answer:
The total frequency=100
The observations are=60,240,300
The relative frequency of bulbs will have a lifespan greater than or equal to 1,000 hours, but less than 2,000 hours=60/600=0.6
In percentages, we can write 0.6*100=60%
The relative frequency of the bulbs will have a lifespan greater than or equal to 2,000 hours, but less than 3,000 hours=240/600=0.4
In percentages, we can write 0.4*100=40%
The relative frequency of the bulbs will have a lifespan greater than or equal to 3,000 hours, but less than 4,000 hours=300/600=0.5
In percentages, we can write 0.5*100=50%

d) If you buy a light bulb from this shipment, what do you predict is the most likely lifespan of your light bulb? Explain your answer using experimental probability.
Answer:
First, compare the life span of the bulbs:
The bulbs will have a lifespan greater than or equal to 1,000 hours, but less than 2,000 hours=60
The bulbs will have a lifespan greater than or equal to 2,000 hours, but less than 3,000 hours=240
The bulbs will have a lifespan greater than or equal to 3,000 hours, but less than 4,000 hours=300
The highest lifespan is 300 so definitely, everyone will predict that it will be the most likely one.
Experimental probability: Experimental probability, also known as Empirical probability, is based on actual experiments and adequate recordings of the happening of events. To determine the occurrence of any event, a series of actual experiments are conducted. Experiments that do not have a fixed result are known as random experiments. The outcome of such experiments is uncertain. Random experiments are repeated multiple times to determine their likelihood. An experiment is repeated a fixed number of times and each repetition is known as a trial. Mathematically, the formula for the experimental probability is defined by;
Probability of an Event P(E) = Number of times an event occurs / Total number of trials.
P(E)=300/600
P(E)=1/2.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 10 Review Test Answer Key

Concepts and Skills

Solve.

Question 1.
You select a card at random from 50 cards numbered from 1 to 50. What are the possible outcomes for the event of choosing a number that is a multiple of 6?
Answer:
The numbers are:1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The total number of outcomes=50
The possible outcomes are 6, 12, 18, 24, 30, 36, 42, 48.
the event is choosing a number that is multiple of 6.
the numbers which are multiple of 6 are 6, 12, 18, 24, 30, 36, 42, 48.

Question 2.
Three fair coins are tossed together once. List the outcomes that are favorable for the event of only two of the coins landing on heads.
Answer:
When 3 coins are tossed, the possible outcomes are HHH, TTT, HTT, THT, TTH, THH, HTH, HHT.
The sample space is S = { HHH, TTT, HTT, THT, TTH, THH, HTH, HHT}
Number of elements in sample space, n(S) = 8
Let E₂ denotes the event of getting two heads.
E₂ = {HHT, HTH, THH}
n(E₂) = 3
P(getting two heads) = n(E₂)/ n(S)
= 3/8
Hence the required probability is 3/8.

Question 3.
Daniel wants to write all the 2-digit numbers with no repeating digits that can be formed using the digits 5, 6, and 7.
a) List all the possible outcomes.
Answer:
The above-given numbers are 5, 6, and 7
The possible numbers will come without repeating digits that can be formed 2-digit numbers=56, 57, 65, 67, 75, 76

b) X is the event that the 2-digit number is divisible by 5. How many of the outcomes are favorable to event X?
Answer: 2 outcomes.
Explanation:
The possible outcomes we got in the above question=56, 57, 65, 67, 75, 76
Event X is the numbers divisible by 5.
The favourable outcomes we need to find.
5*12=65
5*15=75
E(X)=65, 75
Therefore, the number of outcomes is 2

Question 4.
Amy writes a computer program that will choose two letters from her own name to make a two-letter “string.” The order of the letters matters. For example, AM and MA are different strings.
a) List all the possible outcomes for forming a two-letter string.
Answer:
The above-given name: Amy
If we choose the letters A and Y then the possible outcomes for forming a two-letter string are:
AY and YA.

b) What is the probability that Amy forms a two-letter string with the letter M in it?
Answer:
The above-given name: Amy
Now letter M is given
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The outcomes are 2 [AM, MY]
The total number of letters in the name=3
P(A)=2/3
therefore, the probability is 2/3.

Question 5.
Two-digit numbers are formed using digits 2, 3, and 4, with no repeating digits.
a) List all the possible outcomes.
Answer:
The above-given numbers are 2, 3, and 4
The possible numbers will come without repeating digits that can be formed 2-digit numbers=23, 24, 32, 34, 42, 43

b) What is the probability of forming a number greater than 32?
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The numbers we formed using the 2, 3, and 4 are 23, 24, 32, 34, 42, 43
The numbers greater than 32 are 34, 42, 43
the possible outcomes are 3
The total number of outcomes=6
P(greater than 32)=3/6
P(greater than 32)=1/2=0.5
therefore, the probability is 0.5 or 1/2.

Question 6.
Tim has three DVDs. One is a science fiction movie, one is an action movie, and the other is a documentary. If he stacks the DVDs randomly, what is the probability that the science fiction movie is on top, the action movie is in the middle, and the documentary is on the bottom?
Answer:1/3
The above-given data:
The DVDs Tim has science fiction movies, action movies, documentaries.
If he kept randomly on the shelves then the probability of the DVDs keeping in a given order.
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
the total number of DVDs=3
The possible number of outcomes keeping in the DVDs=1
P(science fiction)=1/3
P(action movie)=1/3
P(documentaries)=1/3

Question 7.
A ribbon is selected at random out of a total of 4 orange ribbons, 5 yellow ribbons, and 3 red ribbons. What is the probability of selecting an orange ribbon?
Answer:
The total ribbons=12
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The possible outcomes of orange ribbons=4
P(orange ribbons)=4/12
P(orange ribbons)=1/3
Therefore, the probability of orange ribbons=1/3.

Question 8.
Use the spinner shown.

a) What is the probability of landing on an even number?
Answer:1/2
The above-given numbers 1, 2, 3, and 4
the even numbers are 2 and 4
the possibilities are 2
the total number of outcomes=4
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
P(even number)=2/4
P(even number)=1/2
therefore, the probability of landing an even number is 1/2 or 0.5

b) What is the probability of landing on a number less than 4?
Answer:
The above-given numbers 1, 2, 3, and 4
the numbers less than 4 are 1, 2, 3
the possibilities are 3
the total number of outcomes=4
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
P(less than 4)=3/4
therefore, the probability is 3/4.

Problem Solving

Solve.

Question 9.
Olivia and Jackie played a game with the spinner shown.

Olivia spun a 2 on 12 spins out of 50, while Jackie spun a 2 on 19 spins out of 100.
a) Find each person’s experimental probability of spinning a 2. Express your answers as decimals.
Answer:0.24; 0.19
explanation:
The above-given data
Olivia spun 12 spins on 2 out of 50.
While Jackie spun 19 spins on 2 out of 2
Experimental Probability is the probability of an event based on exact recordings or experiments of an event. The calculation is done by dividing the number of times an event occurred by the total number of trials in an experiment. We can express experimental probability mathematically as,
Experimental Probability = Number of times a particular event occurs/ Number of total trials)
The number of times a particular event occurs for Olivia=12
The number of times a particular event occurs for Jackie=19
The total number of spins of Olivia=50
the total number of spins of Jackie=100
P(Olivia)=12/50
P(Olivia)=6/25
P(Olivia)=0.24
P(Jackie)=19/100
P(Jackie)=0.19

b) Suppose the spinner is fair, meaning that it is equally likely to land on any of the numbers. What is the theoretical probability of spinning a 2?
Answer:1/6
Theoretical Probability describes the probability of the happening of certain events that are not in an experimental way of an occurrence. An example of this is drawing a red stone out of a bag etc. We can express theoretical probability mathematically as,
theoretical Probability = Total number of desired outcomes/ Total number of outcomes
Total number of the desired outcomes=1
Total number of outcomes=6
Theoretical probability=1/6
therefore, the probability of spinning 2 is 1/6.

c) Assuming the spinner is fair, what do you predict will happen to the experimental probability of getting a 2 if the spinner is spun 500 times?
Answer:
It is closer to the theoretical probability.
Theoretical Probability describes the probability of the happening of certain events that are not in an experimental way of an occurrence. An example of this is drawing a red stone out of a bag etc. We can express theoretical probability mathematically as,
theoretical Probability = Total number of desired outcomes/ Total number of outcomes.

Question 10.
A red number die and a green number die each have faces labeled 1 to 6. Suppose you roll the number dice and record the values for each die as an ordered pair of numbers: (red value, green value). The event E is the event of getting a pair of values in which the number on the green die is greater than the number on the red die.
a) Find all the outcomes favorable to event E.
Answer:
The numbers are on the green die=1, 2, 3, 4, 5, 6
The numbers are on the red die=1, 2, 3, 4, 5, 6
The possible outcomes are:
{ (2, 1) (3, 2) (3, 1) (4, 1) (4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (5, 4) (6, 1) (6, 2) (6, 3) (6, 4) (6, 50)}

b) Find P(E).
Answer:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The event E is the event of getting a pair of values in which the number on the green die is greater than the number on the red die.
the number of favourable outcomes=5 (2, 3, 4, 5, 6)
The total number of possible outcomes=15
{ (2, 1) (3, 2) (3, 1) (4, 1) (4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (5, 4) (6, 1) (6, 2) (6, 3) (6, 4) (6, 50)}
P(E)=5/15
P(E)=1/3
In decimals, we can write as 0.3
Therefore, the probability is 0.3 or 1/3.

Question 11.
Two fair counters are tossed together. One of the counters is white on one side and black on the other side, the other counter is white on one side, and red on the other side. They are tossed together and the face up colors of the two counters are noted. Let F be the event that red and black appear, and F be the event that at least one is white.
a) Define the sample space of the experiment.
Answer:
A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments.
Let W, B and R represent white, black and red respectively.
the sample space is {WW, WR, BW, BR}.

b) Calculate the probability of each outcome.
Answer:1/4
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Total number of possible outcomes=4
The number of a favourable outcomes for red=1
The number of a favourable outcome for black=1
The number of a favourable outcome for white=1
P(red)=1/4
P(black)=1/4
P(white)=1/4

c) Construct the probability distribution table. Is it a uniform probability model?
Answer:
A probability distribution table is a table that displays the probability that a random variable takes on certain values.

Uniform in Probability. A discrete uniform probability distribution is one in which all elementary events in the sample space have an equal opportunity of occurring. As a result, for a finite sample space of size n, the probability of an elementary event occurring is 1/ n.
yes, all the outcomes have an equal probability of occurring.

d) Draw a Venn diagram for events E and F. Are they mutually exclusive?
Answer:
A diagram is used to represent all possible relations of different sets. A Venn diagram can be represented by any closed figure, whether it be a Circle or a Polygon (square, hexagon, etc.). But usually, we use circles to represent each set.

In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0
According to the above definition:
Yes, the events are mutually exclusive.

e) Calculate P(E) and P(F).
Answer:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The number of favourable outcomes of event E=1
The total number of possible outcomes=4
P(E)=1/4
The number of favourable outcomes of event F=1
P(F)=3/4

Question 12.
Joan keeps track of the number of emails she receives each day for 100 days. She then makes a table showing how many days she received 0 emails, 1 email, 2 emails, and so on, as shown in the table.

a) Copy and complete the table above by finding each relative frequency.
Answer:
The number of times an event occurs is called a frequency. The number of times an event occurs is called a frequency. Relative frequency is an experimental one, but not a theoretical one. Since it is an experimental one, it is possible to obtain different relative frequencies when we repeat the experiments. To calculate the frequency we need
– Frequency count for the total population
– Frequency count for a subgroup of the population
We can find the relative frequency probability in the following way if we know the above two frequencies. The formula for a subgroup is;
Relative Frequency = Subgroup Count / Total Count

The relative frequency of the number of days(5)=5/100=1/20
The relative frequency of the number of days(20)=20/100=1/5
The relative frequency of the number of days(15)=15/100=3/20
The relative frequency of the number of days(31)=31/100
The relative frequency of the number of days(27)=27/100
The relative frequency of the number of days(22)=2/100=1/50

b) Present the relative frequencies in a bar graph.
Answer:
Relative Frequency Graph Maker A relative frequency graph shows the relative frequencies corresponding to the values in a sample, with respect to the total sample data.

Question 13.
At a music school, 400 students were given a survey on the number of hours they practice each week. The results of this survey are shown in the relative frequency histogram.

a) How many students practiced at least 2 hours but less than 4 hours per week?
Answer: 96
Explanation:
The above-given question:
The total number of students was given a survey on the number of hours they practise each week=400
The number of students practised at least 2 hours but less than 4 hours per week=X
at least 2 means 2 or more than 2.
The percentage of students practised at least 2 hours but less than 4 hours per week=24%
By using the given data we need to calculate the number of students practised between the given time in the question
X=24*400/100
X=96.
Therefore, the number of students practised at least 2 hours but less than 4 hours per week is 96.

b) One of the students is selected at random. What is the probability that this student has practiced 6 or more hours per week?
Answer:0.26 or 13/50
Explanation:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.

The probability that this student has practised 6 or more hours per week=P(A)
The total number of students (total possible outcomes)=n(S)=400
The number of favourable outcomes(6 0r more)=22%+4%=26%
26% of 400
=26*400/100
=104
P(A)=104/400
P(A)=13/50
In decimals, we can write as 0.26
Therefore, the probability is 0.26 or 13/50

Question 14.
Fifty students at a school kept track of how many books they read last semester. Each student wrote the number of books he or she read in a table provided by the school librarian.

a) Copy and complete the frequency table.

Answer:
Frequency table: Frequency means the number of times a value appears in the data. A table can quickly show us how many times each value appears. If the data has many different values, it is easier to use intervals of values to present them in a table.

b) Copy and complete the probability distribution table below.

Answer:
A probability distribution table is a table that displays the probability that a random variable takes on certain values.

The probability of students read books 0-5=15/50=0.3
The probability of students read books 6-10=15/50=0.3
The probability of students reading books 11-15=8/50=0.16
The probability of students reading books 16-20=6/50=0.12
The probability of students reading books 21-25=3/50=0.06
The probability of students reading books 21-25=3/50=0.06

c) Suppose the librarian selects one of the students at random. What is the probability that the student has read at least 6 books but not more than 15 books?
Answer:23/50 or 0.46
Explanation:
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
the probability that the student has read at least 6 books but not more than 15 books=P(A)
The number of outcomes (6-15)=15+8=23=n(E)
the total number of students=50=n(S)
P(A)=23/50
In decimals, we can write as 0.46

d) What per cent of the students had read 10 or fewer books?
Answer: 30%
The above-given question:
The per cent of the students had read 10 or fewer books=X
10 or fewer means 10 and 10 below books.
Now observe the frequency table above:
0-10 there are total 30 books (15+15)
Now calculate the percentage.
X=15/50*100
X=3/10*100
X=30%
therefore, 30% of the students had read 10 or fewer books.

e) Present the probability distribution in a bar graph.
Answer:
Probability distribution yields the possible outcomes for any random event. It is also defined based on the underlying sample space as a set of possible outcomes of any random experiment. These settings could be a set of real numbers or a set of vectors or a set of any entities. It is a part of probability and statistics.

This handy Math in Focus Grade 7 Workbook Answer Key Cumulative Review Chapters 9-10 detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Cumulative Review Chapters 9-10 Answer Key

Concepts and Skills

Find the range, the three quartiles, and the interquartile range. (Lesson 9.1)

Question 1.
32, 65, 90, 25, 46, 81, 30, 57, 85, 104, 33, 48
Answer:
To calculate range follow the below steps:
1. List the elements of your data set.
2. Identify the highest and lowest numbers in the set.
3. Subtract the smallest number in your data set from the largest number.
4. Label the range clearly.
In the given set the highest number is 104
the lowest number is 25
Now subtract highest number-lowest number
104-25=79
therefore, the range is 79.
The three quartiles:
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
the given data: 32, 65, 90, 25, 46, 81, 30, 57, 85, 104, 33, 48
write in ascending order:
25, 30, 32, 33, 46, 48, 57, 65, 81, 85, 90, 104
Now finding the median:
Q2=46, 48, 57, 65
Now add the middle values and divide by 2.
Q2=48+57/2
Q2=105/2
Q2=52.5
Now, look at the bottom half of the data. The median of this half is found between the second and fifth values:
25, 30, 32, 33, 46, 48
Thus the first quartile is found to equal Q₁ = (32 +33)/2
Q1=65/2
Q1=32.5
To find the third quartile, look at the top half of the original data set. We need to find the median of:
57, 65, 81, 85, 90, 104
Here the median is (81 + 85)/2 = 83. Thus the third quartile Q₃ = 83.
Interquartile range=Q3-Q1
Interquartile range=83-32.5
Interquartile range=50.5

Question 2.
12.6, 13.0, 15.5, 18.6, 14.4, 12.0, 11.0, 11.0, 15.6, 15.9
Answer:
11.0, 11.0, 12.0, 12.6, 13.0, 14.4, 15.5, 15.6, 15.9, 18.6
To calculate range follow the below steps:
1. List the elements of your data set.
2. Identify the highest and lowest numbers in the set.
3. Subtract the smallest number in your data set from the largest number.
4. Label the range clearly.
In the given set the highest number is 18.6
the lowest number is 11.0
Now subtract highest number-lowest number
18.6-11.0=7.6
therefore, the range is 7.6
The three quartiles:
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
the given data: 12.6, 13.0, 15.5, 18.6, 14.4, 12.0, 11.0, 11.0, 15.6, 15.9
write in ascending order:
11.0, 11.0, 12.0, 12.6, 13.0, 14.4, 15.5, 15.6, 15.9, 18.6
Now finding the median:
Q2= 13.0, 14.4
Now add the middle values and divide by 2.
Q2=13.0+14.4/2
Q2=27.4/2
Q2=13.7
Now, look at the bottom half of the data. The median of this half is found between the second and fifth values:
11.0, 11.0, 12.0, 12.6, 13.0
Thus the first quartile is found to equal Q₁ = 12.0
Q1=12.0
Note: If the size of the data set is odd, do not include the median when finding the first and third quartiles.
To find the third quartile, look at the top half of the original data set. We need to find the median of:
14.4, 15.5, 15.6, 15.9, 18.6
Here the median is 15.6. Thus the third quartile Q₃ = 15.6.
Interquartile range=Q3-Q1
Interquartile range=15.6-12.0
Interquartile range=3.6

Find the mean absolute deviation. Round to the nearest hundredth. (Lesson 9.3)

Question 3.
103, 111, 150, 165, 192, 144, 144, 163, 121
Answer:
The mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean.
Mean absolute deviation helps us get a sense of how “spread out” the values in a data set are.
Formula to calculate mean absolute deviation:
– Start by finding the mean of the data set.
– Find the absolute value of the difference between each data value and the mean.
– Find the sum of the absolute values and divide the sum by the number of data values.
the given values:
103, 111, 150, 165, 192, 144, 144, 163, 121
We’ll begin by finding the mean of the data set.
Mean: The word average mean middle or central point. Most of the time when we are required to calculate the average of a list of data we are supposed to calculate the ‘mean’ of the given data.
Formula to calculate average (mean) for ungrouped data:
Average is calculated by dividing the sum of the given data by the count of the same data.

n=9
Mean=103+111+150+165+192+144+144+163+121/9
Mean=1293/9
Mean=143.666666667
Mean=144
Then find the absolute value of the difference between each data value and the mean
1. |103-144|=41
2. |111-144|=33
3. |150-144|=6
4. |165-144|=21
5. |192-144|=48
6. |144-144|=0
7. |144-144|=0
8. |163-144|=19
9. |121-144|=23
Lastly, find the sum of the absolute values and divide by the number of the values.
MD=41+33+6+21+48+0+0+19+23/9
MD=191/2
MD=21.222

Question 4.
2.0, 3.2, 4.5, 5.6, 7.0, 7.9, 8.6, 9.1, 10.2, 12.3
Answer:
The mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean.
Mean absolute deviation helps us get a sense of how “spread out” the values in a data set are.
Formula to calculate mean absolute deviation:
– Start by finding the mean of the data set.
– Find the absolute value of the difference between each data value and the mean.
– Find the sum of the absolute values and divide the sum by the number of data values.
the given values:
2.0, 3.2, 4.5, 5.6, 7.0, 7.9, 8.6, 9.1, 10.2, 12.3
We’ll begin by finding the mean of the data set.
Mean: The word average mean middle or central point. Most of the time when we are required to calculate the average of a list of data we are supposed to calculate the ‘mean’ of the given data.
Formula to calculate average (mean) for ungrouped data:
Average is calculated by dividing the sum of the given data by the count of the same data.
Average=X1+X2+…+Xn/n
n=10
Mean=2.0+3.2+4.5+5.6+7.0+7.9+8.6+9.1+10.2+12.3/10
Mean=70.4
Mean=7.04
Mean=7 (rounded to 100)
Then find the absolute value of the difference between each data value and the mean
1. |2.0-7|=5
2. |3.2-7|=3.8
3. |4.5-7|=2.5
4. |5.6-7|=1.4
5. |7.0-7|=0
6. |7.9-7|=0.9
7. |8.6-7|=1.6
8. |9.1-7|=2.1
9. |10.2-7|=3.2
10. |12.3-7|=5.3
Lastly, find the sum of the absolute values and divide by the number of the values.
MD=5+3.8+2.5+1.4+0+0.9+1.6+2.1+3.2+5.3/10
MD=25.8/10
MD=2.58

Use the information below to answer questions 5 to 10. (Lessons 9.1, 9.2)

The heights of plants, in centimeters, are shown in a stem-and-leaf plot.

Question 5.
Find the range of the data.
Answer:
To calculate range follow the below steps:
1. List the elements of your data set.
2. Identify the highest and lowest numbers in the set.
3. Subtract the smallest number in your data set from the largest number.
4. Label the range clearly.
You can use a stem-and-leaf plot to figure out the range of a set of data. The range is the difference between the maximum score and the minimum score.
Let’s look at the above diagram:
The smallest number in the stem-and-leaf plot is 10.5. You can see that by looking at the first stem and the first leaf. The greatest number is the last stem and the last leaf on the chart. In this case, the largest number is 15.9
To find the range, subtract the smallest number from the largest number. This difference will give you the range.
15.9-10.5=5.4 cm
The range is 5.4 for this set of data.

Question 6.
Find the mode of the data.
Answer:
Mode: The mode is the value that appears most often in a set of data. The mode of a discrete probability distribution is the value x at which its probability mass function takes its maximum value. In other words, it is the value that is most likely to be sampled.
1. Now write the numbers in the data set:
{10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6, 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9}
2. Order the numbers from smallest to largest:
{10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6, 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9}
3. Count the number of times each number is repeated:
10.5 – 1; 10.7 – 2; 11.0 – 1; 11.3 – 1; 11.4 – 1; 11.5 – 1; 11.8 – 1; 12.0 – 2; 12.6 – 2; 12.7 – 1; 12.9 – 1; 14.3 – 3; 14.8 – 1; 15.0 – 2;
15.1 – 1; 15.7 – 2; 15.8 – 1; 15.9 – 1}
4. Identify the value(or values) that occur most often:
When you know how many times each value occurs in your data set, find the value that occurs the greatest number of times. This is your data set’s mode. Note that there can be more than one mode in a data set. If the two values are tied for being the most common values in the set, the data set can be said to be bimodal, whereas if three values are tied, the set is trimodal, and so on.
– In the above-given data set, 14.3 occurs more times than any other value, 14.3 is the mode.
– If a value besides 14.3 had also occurred three times, (like, for instance, if there were one more 15.0 in the data set), 14.3 and this other number would both be the mode.

Question 7.
Find the median height.
Answer:
Median: The median is the value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. In simple terms, it may be thought of as the “middle” value of a data set.
The given data set:
{10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6, 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9}
The number of data set=25
Now divide the numbers bottom half and lower half.
{10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6, 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9}
– The bottom half is 12 and the lower half is 12.
– And the median is the middle number that is 12.7
Therefore, 12.7 is the median.

Question 8.
Find the mean height.
Answer:
Mean: The word average mean middle or central point. Most of the time when we are required to calculate the average of a list of data we are supposed to calculate the ‘mean’ of the given data.
Formula to calculate average (mean) for ungrouped data:
Average is calculated by dividing the sum of the given data by the count of the same data.
Average=X1+X2+…+Xn/n
n=25
Average=10.5+10.7+10.7+11.0+11.3+11.4+11.5+11.8+12.0+12.0+12.6+12.6+12.7+12.9+14.3+14.3+14.3+14.8+15.0+15.0+15.1+15.7+15.7+15.8+15.9/25
Average=410.8/25
Average (mean)=16.432

Question 9.
Calculate Q₁ and Q₃.
Answer:
The three quartiles:
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
the given data: {10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6, 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9}
write in ascending order:
{10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6, 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9}
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Data set divided for finding Q1: 10.5, 10.7, 10.7, 11.0, 11.3, 11.4, 11.5, 11.8, 12.0, 12.0, 12.6, 12.6
Q1=11.4+11.5/2
Q1=22.9/2
Q1=11.45
Q2 data set: 12.7, 12.9, 14.3, 14.3, 14.3, 14.8, 15.0, 15.0, 15.1, 15.7, 15.7, 15.8, 15.9
n=13 (odd); If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q3=15.0 cm

Question 10.
Find the interquartile range.
Answer:
The interquartile range IQR is the range in values from the first quartile Q₁ to the third quartile Q₃. Find the IQR by subtracting Q1 from Q3.
IQR=Q3-Q1
IQR=15.0-11.45
IQR=3.55

Use the following information to answer questions 11 to 14. (Lessons 9.1, 9.3)

The history test scores of twenty students are tabulated below.

Question 11.
Find the range of the scores.
Answer:
To calculate range follow the below steps:
1. List the elements of your data set.
2. Identify the highest and lowest numbers in the set.
3. Subtract the smallest number in your data set from the largest number.
4. Label the range clearly.
The highest number is:92
The lowest number is: 50
range=highest number-lowest number
range=92-50
range=42

Question 12.
Find the 3 quartiles of the scores.
Answer:
the given data set: 75, 92, 56, 60, 50, 60, 67, 87, 88, 74, 60, 78, 90, 61, 64, 92, 50, 75, 58, 70.
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
Now write in an ascending order:
50, 50, 56, 58, 60, 60, 60, 61, 64, 67, 70, 74, 75, 75, 78, 87, 88, 90, 92, 92.
Q2=67+70/2
Q2=137/2
Q2=68.5
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Now, look at the bottom half of the data.
50, 50, 56, 58, 60, 60, 60, 61, 64, 67
Q1=60+60/2
Q1=120/2
Q1=60
Thus the first quartile is found to equal Q₁ = 60
Q3 data set: 70, 74, 75, 75, 78, 87, 88, 90, 92, 92
Q3=78+87/2
Q3=165/2
Q3=82.5
Thus the third quartile is found to equal Q3 = 82.5

Question 13.
Find the interquartile range of the scores.
Answer:
The interquartile range IQR is the range in values from the first quartile Q₁ to the third quartile Q₃. Find the IQR by subtracting Q1 from Q3.
IQR=Q3-Q1
IQR=82.5-60
IQR=22.5

Question 14.
Find the mean absolute deviation of the scores. Round to the nearest hundredth.
Answer:
The mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean.
Mean absolute deviation helps us get a sense of how “spread out” the values in a data set are.
Formula to calculate mean absolute deviation:
– Start by finding the mean of the data set.
– Find the absolute value of the difference between each data value and the mean.
– Find the sum of the absolute values and divide the sum by the number of data values.
the given values:
50, 50, 56, 58, 60, 60, 60, 61, 64, 67, 70, 74, 75, 75, 78, 87, 88, 90, 92, 92.
We’ll begin by finding the mean of the data set.
Mean: The word average mean middle or central point. Most of the time when we are required to calculate the average of a list of data we are supposed to calculate the ‘mean’ of the given data.
Formula to calculate average (mean) for ungrouped data:
Average is calculated by dividing the sum of the given data by the count of the same data.
Average=x1+x2+…+x3/n
n=20
Mean=50+50+56+58+60+60+60+61+64+67+70+74+75+75+78+87+88+90+92+92/20
Mean=1407/20
Mean=70.35
Mean=70 (round to 100)
Then find the absolute value of the difference between each data value and the mean
|50-70| = 20
|50-70| = 20
|56-70| = 14
|58-70| = 12
|60-70| = 10
|60-70| = 10
|60-70| = 10
|61-70| = 9
|64-70| = 6
|67-70| = 3
|70-70| = 0
|74-70| = 4
|75-70| = 5
|75-70| = 5
|78-70| = 8
|87-70| = 17
|88-70| = 18
|90-70| = 20
|92-70| = 22
|92-70| = 22
Lastly, find the sum of the absolute values and divide by the number of the values.
MD=20+20+14+12+10+10+10+9+6+3+0+4+5+5+8+17+18+20+22+22/20
MD=235/20
MD=11.75

Use the data in the table to answer questions 15 to 17. (Lesson 9.3)

The table shows the speeds, ¡n miles per hour, of twelve vehicles.

Question 15.
Calculate Q₁, Q₂, and Q₃.
Answer:
The given data set: 80, 65, 72, 58, 60, 70, 75, 68, 48, 51, 88, 90
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
Now write in ascending order:
48, 51, 58, 60, 65, 68, 70, 72, 75, 80, 88, 90.
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=68+70/2
Q2=138/2
Q2=69
Now calculate Q1 and Q3
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 48, 51, 58, 60, 65, 68
Q1=58+60/2
Q1=118/2
Q1=59
Thus, the first quartile is found to equal Q₁ = 59
Q3 data set: 70, 72, 75, 80, 88, 90
Q3=75+80/2
Q3=155/2
Q3=77.5
Thus, the third quartile is found to equal Q3 = 77.5

Question 16.
Draw a box plot of the speeds of the vehicles.
Answer:
The given data set: 80, 65, 72, 58, 60, 70, 75, 68, 48, 51, 88, 90
Now write in ascending order:
48, 51, 58, 60, 65, 68, 70, 72, 75, 80, 88, 90.
Q1=59, Q2=60, Q3=77.5

Question 17.
Calculate the mean absolute deviation of the speeds of vehicles. Round to the nearest hundredth.
Answer:
The mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean.
Mean absolute deviation helps us get a sense of how “spread out” the values in a data set are.
Formula to calculate mean absolute deviation:
– Start by finding the mean of the data set.
– Find the absolute value of the difference between each data value and the mean.
– Find the sum of the absolute values and divide the sum by the number of data values.
the given values:
80, 65, 72, 58, 60, 70, 75, 68, 48, 51, 88, 90
We’ll begin by finding the mean of the data set.
Mean: The word average mean middle or central point. Most of the time when we are required to calculate the average of a list of data we are supposed to calculate the ‘mean’ of the given data.
Formula to calculate average (mean) for ungrouped data:
Average is calculated by dividing the sum of the given data by the count of the same data.
Average=x1+x2+…+x3/n
n=12
Mean=80+65+72+58+60+70+75+68+48+51+88+90/12
Mean=825/12
Mean=68.75
Mean=69 (round to nearest 100)
Then find the absolute value of the difference between each data value and the mean
|80-69| = 11
|65-69| = 4
|72-69| = 3
|58-69| = 11
|60-69| = 9
|70-69| = 1
|75-69| = 6
|68-69| = 1
|48-69| = 21
|51-69| = 18
|88-69| = 19
|90-69| = 21
Lastly, find the sum of the absolute values and divide by the number of the values.
MD=11+4+3+11+9+1+6+1+21+18+19+21/12
MD=125/12
MD=10.41

Use the box plot to answer questions 18 to 20 (Lesson 9.3)

The box plot below summarizes the reaction times, in seconds, of 300 drivers in an experiment.

Question 18.
State the 5-point summary.
Answer:
– When we display the data distribution in a standardized way using 5 summary – minimum, Q1 (First Quartile), median, Q3(third Quartile), and maximum, it is called a Box plot.
– The method to summarize a set of data that is measured using an interval scale is called a box and whisker plot. These are the maximum used for data analysis.
– It does not show the distribution in particular as much as a stem and leaf plot or histogram does.
– In simple words, we can define the box plot in terms of descriptive statistics related concepts. That means box or whiskers plot is a method used for depicting groups of numerical data through their quartiles graphically.
– These may also have some lines extending from the boxes or whiskers which indicates the variability outside the lower and upper quartiles, hence the terms box-and-whisker plot and box-and-whisker diagram. Outliers can be indicated as individual points.
– It helps to find out how much the data values vary or spread out with the help of graphs. As we need more information than just knowing the measures of central tendency, this is where the box plot helps. This also takes less space. It is also a type of pictorial representation of data.
From the above-given box plot, we need to summarize the points.
according to the definition:
The first point represents the minimum, the point is 1.5
The second point represents Q1 (First Quartile), the Q1 point is 2.0
The third point represents Q2 (Median), the Q2 point is 2.5
The fourth point represents Q3 (third quartile), the Q3 point is 2.7
The fifth point represents the maximum, Thus the maximum point is 3.1

Question 19.
How many drivers have reaction time more than 2 seconds?
Answer: 225 drivers.

Question 20.
If a driver with reaction time less than 2 seconds is considered fast, how many drivers are fast?
Answer:

Solve. Show your work. (Lesson 9.5)

Question 21.
Six random samples of inner diameters, in millimeters, of metal pipes were collected. The sample means are 112, 103.5, 98.4, 106.2, 110, and 99.7. Use the 6 sample means to generate a mean length to estimate the population mean diameter of the pipes. Round your answer to the nearest tenth.
Answer:
The given sample means are 112, 103.5, 98.4, 106.2, 110, and 99.7
Mean length=112+103.5+98.4+106.2+110+99.7/6
Mean length=629.8/6
Mean length=104.96
Therefore, mean length=105 mm (round to nearest 10).

Question 22.
A bag contains 5 red cards, 4 yellow cards, and 7 blue cards. A card is randomly drawn from the bag. (Lessons 10.1, 10.2)
a) How many outcomes are in the sample space?
Answer:
The total number of cards=16 cards.
The number of red cards=5
The number of yellow cards=4
The number of blue cards=7
A sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.
The outcomes are {R, R, R, R, R, Y, Y, Y, Y, B, B, B, B, B, B, B}
the number of possibles is 16

b) If X is the event of drawing a card which is not yellow, what are the outcomes favorable to X?
Answer:
If not drawing yellow means then there is the chance of taking blue and red
The possible outcome of event x is { R, R, R, R, R, B, B, B, B, B, B, B}

c) Find P(X).
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The possible outcome of event x is { R, R, R, R, R, B, B, B, B, B, B, B}
the total number of possible outcomes=16
the favourable outcomes=12
P(X)=n(E)/n(S)
P(X)=12/16
P(X)=3/4
P(X)=0.75
Therefore, the probability of event X is 0.75

d) If Y is the event of drawing a blue card, what does the complement of Y mean?
Answer:
If the probability of occurring an event is P(A) then the probability of not occurring an event is
P(A’) = 1- P(A)
P(Y)=7/16
The number of favourable outcomes is 7
We need to find out the not occurring of an event.
The complement of event Y is:
P(Y’)=1-P(Y)
P(Y’)=1-7/16
P(Y’)=16-7/16
P(Y’)=9/16
P(Y’)=0.5

Question 23.
The numbers in the Venn diagram indicate the number of outcomes favorable to the events. For example, there are 12 outcomes favorable to event A. (Lesson 10.2)

a) What is the total number of possible outcomes?
Answer: 48
Explanation:
The number of outcomes to event A=12
The number of outcomes to event B=21
the number of outcomes that are not event A and event B=15
Now add all the outcomes=12+21+15
The total number of possible outcomes are 48.

b) Find P(A), P(B), and P(B’).
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The total number of possible outcomes=48
The number of favourable outcomes of event A=12
P(A)=12/48
P(A)=1/4
The number of favourable outcomes of event B=21
P(B)=21/48
P(B)=7/16
If the probability of occurring an event is P(B) then the probability of not occurring an event is
P(B’) = 1- P( B)
Here we know P(B) is 7/16, then we can find the complement of B.
P(B’)=1-7/16
P(Y’)=16-7/16
P(Y’)=9/16
P(Y’)=0.5

c) Copy and complete the table.

Answer:
The number of outcomes to event A=12
The number of outcomes to event B=21
the number of outcomes that are not event A and event B=15
Now add all the outcomes=12+21+15
The total number of possible outcomes is 48.
Definition of Mutually Non-Exclusive Events: Two events A and B are said to be mutually non-exclusive events if both events A and B have at least one common outcome between them. The event of getting an ‘odd-face’ and the event of getting ‘less than 4’ occur when we get either 1 or 3.
In logic and probability theory, two events (or propositions) are mutually exclusive or disjoint if they cannot both occur at the same time. A clear example is the set of outcomes of a single coin toss, which can result in either heads or tails, but not both.
according to the above definitions we can fill the table:

Question 24.
A coin is flipped repeatedly and the outcomes are HTTHHHTHTHTHTHHTHHTH. What is the experimental probability that the coin lands on heads when you flip the coin again? (Lesson 10.3)
Answer:
The total outcomes came when the coin was flipped repeatedly=20
The number of times coin lands on the head when the coin is flipped again=12
Experimental Probability is the probability of an event based on exact recordings or experiments of an event. The calculation is done by dividing the number of times an event occurred by the total number of trials in an experiment. We can express experimental probability mathematically as,
experimental probability=Number of times a particular event occurs/Number of total trails.
P(Heads)=12/20
P(Heads)=3/5
P(Heads0=0.6
Therefore, the experimental probability that the coin lands on heads when you flip the coin again is 0.6 or 3/5

Question 25.
A spinner with outcomes A, E, O, and B is spun and the letter at which the arrow is pointing is recorded. After spinning the spinner several times, it generates the following outcomes: OABBBEAOOBAA. What is the experimental probability that the arrow points at a vowel if you spin the spinner again? (Lesson 10.3)
Answer:
Experimental Probability is the probability of an event based on exact recordings or experiments of an event. The calculation is done by dividing the number of times an event occurred by the total number of trials in an experiment. We can express experimental probability mathematically as
experimental probability=Number of times a particular event occurs/Number of total trails.
The total number of outcomes=12
The number of times a particular event occurs=8 (favourable outcomes of vowels)
According to the definition:
P(Vowels)=8/12
P(Vowels)=2/3

Problem Solving

Solve. Show your work.

Question 26.
The table shows 100 responses on the level of satisfaction of a service. Customers evaluate their experiences, as 1 very dissatisfied, 2 somewhat dissatisfied, 3 neutral, 4 somewhat satisfied or 5 very satisfied. (Chapters 9, 10)

a) Find the lower quartile, median, and upper quartile.
Answer:
the given data set: 7, 18, 27, 23, 25.
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
Now write in ascending order:
7, 18, 23, 25, 27
Q2=23
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Now, look at the bottom half of the data.
Q1=7+8/2
Q1=25/2
Q1=12.5
Thus the first quartile is found to equal Q₁ = 12.5
Q3=25+27/2
Q3=52/2
Q3=26
Thus the third quartile is found to equal Q3 = 26

b) Draw a box plot of the level of satisfaction.
Answer:
– When we display the data distribution in a standardized way using 5 summary – minimum, Q1 (First Quartile), median, Q3(third Quartile), and maximum, it is called a Box plot.
– The method to summarize a set of data that is measured using an interval scale is called a box and whisker plot. These are the maximum used for data analysis.

c) If a customer’s response is randomly selected, what is the probability that the level of satisfaction is at least 4?
Answer:
The frequency of 4 and 5 are 23 and 25
The favourable outcomes are 2
The total outcomes are 23+25=48
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
P(satisfaction)=2/48
P(satisfaction)=1/24
In decimals, we can write as 0.04
Therefore, the probability that the level of satisfaction is at least 4 is 1/24 or 0.04

d) Construct a relative frequency table. Write each relative frequency as a percent.
Answer:
The number of times an event occurs is called a frequency. Relative frequency is an experimental one, but not a theoretical one. Since it is an experimental one, it is possible to obtain different relative frequencies when we repeat the experiments. To calculate the frequency we need
– Frequency count for the total population
– Frequency count for a subgroup of the population
We can find the relative frequency probability in the following way if we know the above two frequencies. The formula for a subgroup is;
Relative Frequency = Subgroup Count / Total Count
The relative frequency of very dissatisfied=1/7
The relative frequency of dissatisfied=2/18=1/9
The relative frequency of neutral=3/27=1/9
The relative frequency of somewhat satisfied=4/23
the relative frequency of very satisfied=5/25=1/5

e) Present the relative frequency in a bar graph.
Answer:
The relative frequency of very dissatisfied=1/7=0.14
The relative frequency of dissatisfied=2/18=1/9=0.11
The relative frequency of neutral=3/27=1/9=0.11
The relative frequency of somewhat satisfied=4/23=0.17
the relative frequency of very satisfied=5/25=1/5=0.2

Question 27.
A survey was conducted on 100 randomly selected people about the types of books they usually read. The survey results show that 58 people read novels, 40 people read science fiction, and 10 people read neither. (Chapter 10)
a) Find the number of people surveyed who read both novels and science fiction.
Answer:8
The number of people who read novels is 58
The number of people who reads science fiction is 40
The number of people read none=10
Therefore, 100-10=90
The remaining people are 90.
If we subtract science fiction people from the remaining people then we get the people who read novels.
90-40=50
Thus, 50 people read novels.
But we have 58 people read the novel
: 58-50=8
8 people read both novels and science fiction.
So, a number of people who read only science fiction: 40-8=32.

b) Draw a Venn diagram to represent the different types of books read.
Answer:
A diagram used to represent all possible relations of different sets. A Venn diagram can be represented by any closed figure, whether it be a Circle or a Polygon (square, hexagon, etc.). But usually, we use circles to represent each set.

c) Copy and complete the following relative frequency table. Write each relative frequency as a decimal.

Answer:
relative frequency: We come across repetitive data in observation all the time. When repetitive data is given, then the number of times it has been in the observation is called its frequency. Relative frequency is the comparison between the number of times a number has been repeated to the total frequencies of all the numbers. Mathematically speaking, relative frequency is the division between the individual frequency of an item by the total number of repetition that has occurred.
The formula for the relative frequency is given as:
relative frequency=f/n
Here,
f is the number of times the data occurred in an observation
n  = total frequencies
The relative frequency of reading novel=50/100=0.5
The relative frequency of science fiction=32/100=0.32
The relative frequency of both novel and science fiction=8/100=0.08
The relative frequency of reading neither=10/100=0.1

d) Draw a relative frequency bar graph using decimals.
Answer:
The line plot is a useful graph for examining small sets of data. It’s especially helpful as a device for learning basic statistical ideas. But for larger data sets, it can be awkward to create, since for each data value there is a corresponding dot. That’s a lot of dots for data sets with hundreds or thousands of values! You can, however, replace a line plot with a frequency bar graph.

e) If a person is randomly selected from this population to do the survey, what is the probability that the selected person reads novels?
Answer: 0.58
Explanation:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
the number of people who read novels=58
The total number of people=100
P(novels)=58/100
P(novels)=0.58
Therefore, the probability is o.58

Question 28.
The table shows the number of children surveyed for each age group. (Chapter 10)

a) Construct a probability model. Express each probability of the age groups as a percent.
Answer:
A probability model is a mathematical representation of a random phenomenon. It is defined by its sample space, events within the sample space, and probabilities associated with each event. The sample space S for a probability model is the set of all possible outcomes.
The probability 0f age group 10 to 12=14/80=0.175=17.5%
The probability of age group 12 to 14=28/80=0.35=35%
The probability of age group 14 to 16=20/80=0.25=25%
The probability of age group 16 to 18=18/80=0.225=22.5%

b) If a child is randomly selected, what is the probability that the selected child is at least 10 years old but not yet 14 years old?
Answer:
The probability 0f age group 10 to 12=14
The probability 0f age group 12 to 14=28
here asked the question that we need to find out the probability that the selected child is at least 10 years old but not yet 14 years old.
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The number of favourable outcomes is 14+28=42
The total number of possible outcomes=80
P(age)=42/80
P(age)=0.525
Therefore, the probability is 0.525

c) Display the probability distribution in a histogram using percent.
Answer:
Probability distribution yields the possible outcomes for any random event. It is also defined based on the underlying sample space as a set of possible outcomes of any random experiment. These settings could be a set of real numbers or a set of vectors or a set of any entities. It is a part of probability and statistics.

Question 29.
The data show the masses, in kilograms, of 14 grass carps. (Chapters 9, 10)

a) Construct a stem-and-leaf plot of the masses of the grass carps.
Answer:
Definition: The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making stem-and-leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.
By using the above-given data, we can make a stem-and-leaf plot:

b) Make a frequency table. Group the data into 4 intervals: 5-10, 10-15, 15-20, and 20 — 25. Note: The interval 5 — 10 includes masses greater than or equal to 5 kilograms, but less than 10 kilograms.
Answer:
Frequency refers to the number of times an event or a value occurs.  A frequency table is a table that lists items and shows the number of times the items occur. We represent the frequency by the English alphabet ‘f’.
Creating a frequency table:
Step 1: Make three columns. The first column carries the data values in ascending order (from lesser to large values).
Step 2: The second column contains the number of times the data value occurs using tally marks. Count for every row in the table. Use tally marks for counting.
Step 3:  Count the number of tally marks for each data value and write it in the third column.

c) Draw a probability model. Give probabilities in fractions.
Answer:
A probability model is a mathematical representation of a random phenomenon. It is defined by its sample space, events within the sample space, and probabilities associated with each event. The sample space S for a probability model is the set of all possible outcomes.
– Identify every outcome.
– Determine the total number of possible outcomes.
– Compare each outcome to the total number of possible outcomes.
The probability of mass 5-10 is 4/14=2/7
The probability of mass 10-15 is 3/14
The probability of mass 15-20 is 4/14=2/7
The probability of mass 20-25 is 3/14

d) Present the probability distribution, in fractions, in a histogram.
Answer:
Probability distribution yields the possible outcomes for any random event. It is also defined based on the underlying sample space as a set of possible outcomes of any random experiment. These settings could be a set of real numbers or a set of vectors or a set of any entities. It is a part of probability and statistics.

e) Is the model a uniform probability model? Explain your answer.
Answer: No, all possible outcomes do not have the same relative frequency.
A discrete uniform probability distribution is one in which all elementary events in the sample space have an equal opportunity of occurring. As a result, for a finite sample space of size n, the probability of an elementary event occurring is 1/n. Uniform distributions are very common for initial studies of probability. The histogram of this distribution will look rectangular in shape.

Question 30.
In a game, the scores given to each student are in multiples of 5 up to a maximum of 25. The bar graph and dot plot show the scores of Groups A and B, respectively. (Chapters 9, 10)

a) Calculate the mean score of each group.
Answer:
Mean: The word average mean middle or central point. Most of the time when we are required to calculate the average of a list of data we are supposed to calculate the ‘mean’ of the given data.
Formula to calculate average (mean) for ungrouped data:
Average is calculated by dividing the sum of the given data by the count of the same data.
Average=X1+X2+…+Xn/n
n=5
Finding mean for Group A:
The given data is 2, 5, 8, 4, 1
Mean=2+5+8+4+1/5
Mean=20/5
Mean of group A=4
Finding mean for group B:
The given data is 4, 7, 2, 4, 3
Mean=4+7+2+4+3/5
mean=20/5
Mean of group B=4

b) Calculate the mean absolute deviation of the scores of each group.
Answer:
The mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean.
Mean absolute deviation helps us get a sense of how “spread out” the values in a data set are.
Formula to calculate mean absolute deviation:
– Start by finding the mean of the data set.
– Find the absolute value of the difference between each data value and the mean.
– Find the sum of the absolute values and divide the sum by the number of data values.
the given values of group A: 2, 5, 8, 4, 1
Mean=4
|2 – 4|=2
|5 – 4|=1
|8 – 4|=4
|4 – 4|=0
|1 – 4|=3
MAD=2+1+4+0+3/5
MAD=10/5
MAD of group A=2
The given values of group B: 4, 7, 2, 4, 3
Mean=4
|4 – 4|=0
|7 – 4|=3|
|2 – 4|=2
|4 – 4|=0
|3 – 4|=1
MAD=0+3+2+0+1/5
MAD=6/5
MAD of group B is 1.2

c) Which group’s scores deviate more about its mean score?
Answer:
The mean of group A is 4
The mean of group B is 4
The mean deviation of group A is 2
The mean deviation of Group B is 1.2
By comparing all these Group A score deviate is more.

d) A randomly selected student from one of these groups has a score of 10 or 15. Which group is the student likely to be selected from? Explain your answer using probability.
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
The number of students of 10 or 15 in group A is 5+8=13
the number of students 0f 10 or 15 in group B is 7+2=9
The favourable outcomes of group A are 13
The total number of students=20
P(Group A)=13/20
The favourable outcomes of group B are 9
The total number of students =20
P(Group B)=9/20

Question 31.
The table shows the place of origin of incoming freshman students at a university. (Chapter 9)

Describe how you would conduct a stratified random sampling of 500 of the 6,000 incoming freshman, considering the student distribution within their place of origins.
Answer:
Use the per cent to determine how many of the 500-student samples will be selected from each region. Then conduct a simple random sampling within each region. The total number of students sampled will be 500.

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 12 Practice 1 Counting to 40 to score better marks in the exam.

Math in Focus Grade 1 Chapter 12 Practice 1 Answer Key Counting to 40

First make tens, then count on. Fill in the missing numbers.

Example

Question 1.

____, . . . 20, . . . _____, 31, ____, ______
_______
Answer: 10, . . . 20, . . . 30, 31, 32, 33,34

Question 2.

10, . . . ____, . . . ____, ____, _____
33, ____, _____, 36
Answer:10, . . .20, . . . 30,31, 32,33, 34, 35, 36

Circle groups of 10. Then count and write the numbers.

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Count the s and write the numbers.

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Write the numbers.

Question 12.
twenty-five _______
Answer: twenty-five in numeric is 25

Question 13.
thirty-nine ____
Answer: thirty-nine in numeric is 39

Question 14.
thirty-two ____
Answer: thirty-two in numeric is 32

Question 15.
twenty-nine ____
Answer: twenty-nine in numeric is 29

Write the numbers in words.

Question 16.
21 _______________
Answer: 21 in words is twenty one

Question 17.
37 ________________
Answer: 37 in words is thirty seven

Question 18.
22 ____________
Answer: 22 in words is twenty two

Question 19.
40 _______________
Answer:40 in words is forty

Question 20.
35 _______________
Answer: 35 in words is thirty five

Question 21.
31 _______________
Answer: 31 in words is thirty one

Fill in the missing numbers.

Question 22.
20 + 3 = ____
Answer: 20 + 3 = 23

Question 23.
8 + 30 = ____
Answer: 8 + 30 = 38

Question 24.
_____ + 9 = 39
Answer: 30 + 9 = 39

Question 25.
30 and 5 make ___.
Answer: 30 and 5 make 35

Question 26.
7 and 20 make ___.
Answer: 7 and 20 make 27

Question 27.
___ and 2 make 32.
Answer: 30 and 2 make 32.

Question 28.
___ and 8 make 28.
Answer: 20 and 8 make 28.

Question 29.
___ and 6 make 36.
Answer: 30 and 6 make 36.

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 12 Numbers to 40 to score better marks in the exam.

Math in Focus Grade 1 Chapter 12 Answer Key Numbers to 40

Put On Your Thinking Cap!

Challenging Practice

Fill in the blanks.

Wayne has four cards. Each card has a number on it.

Question 1.
Use two cards to form the least number. Do not begin with 0.
____________________
Answer: card number 1 and card number 2 makes a value 12 which is the least number when not starting with 0

Question 2.
Use two cards to form the greatest number.
____________________
Answer: card number 3 and card number 2 makes a value 32 which is the greatest number

Question 3.
Use two cards to form a number less than 30. Do not begin with 0.
____________________
Answer: card 2 and card 3 will make a number 23 which is a number that is less than 30 and which does not start with 0

Question 4.
Use two cards to form a number greater than 25.
____________________
Answer: card number 3 and card number 2 makes a value 32 which is the greatest number when compares to 25

There is more than one correct answer for Exercises 3 and 4

Put on your Thinking Cap!

Problem Solving

Fill in the blanks.

Jan uses tiles to make shapes that form a pattern.

Answer:

Question 1.
Draw Shape 4.
Answer:

Question 2.
Jan needs _______ more tiles to make Shape 4.
Answer: Jan needs four more tiles to make Shape 4.

Question 3.
Complete the table.

Answer:

Question 4.
Complete the number pattern.
1, 5, 9, ___, ___, _____
Answer: 1, 5, 9, 13, 17, 21

Chapter Review/Test

Vocabulary

Match.

Question 1.

Answer:

Concepts and Skills

Count to find how many.

Question 2.

Answer:

Question 3.

Answer:

Write the numbers.

Question 4.
twenty-four _____
Answer: twenty-four is 24

Question 5.
thirty _____
Answer: thirty is 30

Write the number that ¡s greater.

Question 6.

Answer:

Question 7.

Answer:

Write the number that is less.

Question 8.

Answer:

Question 9.

Answer:

Compare the numbers. Then fill in the blanks.

Question 10.
Which number is greatest? _____
Answer: 35 is the greatest when compared to all the rest

Question 11.
Which number is least? ________
Answer: 9 is the lest when compares to all the other numbers

Order the numbers from greatest to least.

Question 12.

Answer:

Complete the number pattern.

Question 13.
17, 20, 23, ___, ___, 32
Answer:
17,20,23,26,29,32

Fill in the blanks.

Question 14.
2 less than 25 is ____.
Answer: 2 less than 25 is 23.

Question 15.
________ is 3 more than 18.
Answer: 21 is 3 more than 18.

Question 16.
24 = ________ tens 4 ones
Answer: 24 = 2 tens 4 ones

Question 17.
18 = 1 ten _______ ones
Answer: 18 = 1 ten 8 ones

Practice the problems of Math in Focus Grade 4 Workbook Answer Key Chapter 7 Practice 3 Comparing Decimals to score better marks in the exam.

Math in Focus Grade 4 Chapter 7 Practice 3 Answer Key Comparing Decimals

Use the number line. Find the number that is described.

Question 1.
0.1 more than 0.2. _________
Answer:

Explanation:
0.1 more than 0.2 = 0.1 + 0.2 = 0.3.

Question 2.
0.3 more than 0.5. ________
Answer:

Explanation:
0.3 more than 0.5 = 0.3 + 0.5 = 0.8.

Question 3.
0.1 less than 0.6. ___________
Answer:

Explanation:
0.1 less than 0.6 = 0.6 – 0.1 = 0.5.

Question 4.
0.2 less than 0.9. __________
Answer:

Explanation:
0.2 less than 0.9 = 0.9 – 0.2 = 0.7.

Use the number line. Find the number that is described.

Question 5.
0.01 more than 0.13. __________
Answer:

Explanation:
0.01 more than 0.13 = 0.01 + 0.13 = 0.14.

Question 6.
0.04 more than 0.16. ______
Answer:

Explanation:
0.04 more than 0.16 = 0.04 + 0.16 = 0.20.

Question 7.
0.01 less than 0.18. _________
Answer:

Explanation:
0.01 less than 0.18 = 0.18 – 0.01 = 0.17.

Question 8.
0.05 less than 0.17. _______
Answer:

Explanation:
0.05 less than 0.17 = 0.17 – 0.05 = 0.12.

Fill in the missing numbers.
Question 9.

Answer:

Explanation:
0.1 more than 4.7 = 0.1 + 4.7 = 4.8.
0.1 less than 4.7 = 4.7 – 0.1 = 4.6.

Question 10.

Answer:

Explanation:
0.1 more than 2.05 = 0.1 + 2.05 = 2.15.
0.1 less than 2.05 = 2.05 – 0.1 = 1.95.

Question 11.

Answer:

Explanation:
0.01 more than 0.94 = 0.01 + 0.94 = 0.95.
0.01 less than 0.94 = 0.94 – 0.01 = 0.93.

Question 12.

Answer:

Explanation:
0.01 more than 3.8 = 0.01 + 3.8 = 3.81.
0.01 less than 3.8 = 3.8 – 0.01 = 3.79.

Complete the number patterns. Use the number line to help you.

Question 13.
0.2 0.4 0.6 _________ _________
Answer:
0.2     0.4    0.6   0.8   1.0

Explanation:
0.2     0.4    0.6   0.8   1.0
Difference: 0.4 – 0.2 = 0.2.
0.2 + 0.2 = 0.4.
0.4 + 0.2 = 0.6.
0.6 + 0.2 = 0.8.
0.8 + 0.2 = 1.0

Question 14.
1.1 0.9 0.7 _________ ___
Answer:
1.1    0.9    0.7    0.5     0.3

Explanation:
1.1    0.9    0.7    0.5     0.3
Difference: 1.1 – 0.9 = 0.2.
1.1 – 0.2 = 0.9.
0.9 – 0.2 = 0.7.
0.7 – 0.2 = 0.5.
0.5 – 0.2 = 0.3.

Question 15.
0.1 0.4 ________ 1.0 _______
Answer:
0.1    0.4     0.7     1.0     1.3.

Explanation:
0.1    0.4     0.7     1.0     1.3.
Difference: 0.4 – 0.1 = 0.3.
0.1 + 0.3 = 0.4.
0.4 + 0.3 = 0.7.
0.7 + 0.3 = 1.0.
1.0 + 0.3 = 1.3.

Question 16.
2.0 _________ _________ 0.8 0.4
Answer:
2.0      1.6      1.2     0.8      0.4

Explanation:
2.0      1.6      1.2     0.8      0.4
Difference: 0.8 – 0.4 = 0.4.
2.0 – 0.4 = 1.6.
1.6 – 0.4 = 1.2.
1.2 – 0.4 = 0.8.
0.8 – 0.4 = 0.4.

Continue the number patterns.
Question 17.

Answer:
0.03     0.06    0.09     0.12     0.15

Explanation:
0.03     0.06    0.09     0.12     0.15
Difference: 0.06 – 0.03 = 0.03.
0.03 + 0.03 = 0.06.
0.06 + 0.03 = 0.09.
0.09 + 0.03 = 0.12.
0.12 + 0.03 = 0.15.

Question 18.

Answer:
0.24     0.20     0.16      0.12      0.08

Explanation:
0.24     0.20     0.16      0.12      0.08
Difference: 0.24 – 0.20 = 0.04.
0.24 – 0.04 = 0.20.
0.20 – 0.04 = 0.16.
0.16 – 0.04 = 0.12.
0.12 – 0.04 = 0.08.

Practice the problems of Math in Focus Grade 4 Workbook Answer Key Chapter 8 Practice 4 Real-World Problems: Decimals to score better marks in the exam.

Math in Focus Grade 4 Chapter 8 Practice 4 Answer Key Real-World Problems: Decimals

Solve. Show your work.

Example
1 pound of grapes costs $1.79 and 1 pound of peaches costs $1.49. What is the total cost of 1 pound of grapes and 1 pound of peaches?
Cost of grapes + cost of peaches
$1,79 + $1,49 = $3,20
The total cost of 1 pound of grapes and 1 pound of peaches is $323.

Question 1.
A tank is full of water. After 16.5 liters of water are used, 8.75 liters of water are left. How much water was in the full tank?
Answer:
Number of liters of water was in the full tank = 25.25.

Explanation:
Number of liters of water are used = 16.5.
Number of liters of water are left = 8.75.
Number of liters of water was in the full tank = Number of liters of water are used  +  Number of liters of water are left
= 16.5 + 8.75
= 25.25.

Question 2.
A piece of fabric is 4.5 yards long. A customer buys 2.35 yards of the fabric. How many yards of fabric are left?
Answer:
Number of yards of  fabric are left = 2.15.

Explanation:
Number of yards of a piece of fabric = 4.5 yards.
Number of yards of the fabric a customer buys = 2.35 yards.
Number of yards of  fabric are left = Number of yards of a piece of fabric – Number of yards of the fabric a customer buys
= 4.5 – 2.35
= 2.15.

Question 3.
Mr. Larson lives 8.7 miles from school. He was driving home from school and stopped 3.5 miles along the way at a supermarket. How much farther does he have to drive before he reaches home?
Answer:
Number of miles he have to drive before he reaches home = 5.2.

Explanation:
Number of miles Mr. Larson lives from school = 8.7.
Number of miles he was driving home from school and stopped along the way at a supermarket = = 3.5.
Number of miles he have to drive before he reaches home = Number of miles Mr. Larson lives from school – Number of miles he was driving home from school and stopped along the way at a supermarket
= 8.7 – 3.5
= 5.2.

Question 4.
A grocery store is having a sale. A loaf of wheat bread regularly costs $2.29, but the sale price is $1.79. The store is also offering 50¢ off on a gallon of fresh milk. If Mrs. Larson buys a gallon of fresh milk and a loaf of wheat bread, how much does she save on her purchases?
Answer:
Amount of money she save on her purchases = $1.0.

Explanation:
Cost price of a loaf of wheat bread regularly = $2.29.
Selling price of the loaf of wheat bread = $1.79.
Cost of fresh milk a store offering = 50¢.
Amount of money she save on her purchases = (Cost price of a loaf of wheat bread regularly – Selling price of the loaf of wheat bread ) + Cost of fresh milk a store offering
= ($2.29 – $1.79) + 50¢
= $0.5 + 50¢
= $1.0

Question 5.
Lily bought a skirt for $25.90 and a shirt for $19.50. She paid the cashier $50. How much change did she receive?
Answer:
Amount of change she received = $4.6.

Explanation:
Cost of a skirt Lily bought = $25.90.
Cost of a shirt Lily bought = $19.50.
Amount of money she paid the cashier = $50.
Amount of change she received = Amount of money she paid the cashier – (Cost of a skirt Lily bought  + Cost of a shirt Lily bought)
= $50 – ( $25.90 + $19.50)
= $50 – $45.40
= $4.6.

Question 6.
Shannon collects rainwater to water her flowers. She has one bucket with 3.4 gallons of water and another with 1.85 gallons less. She uses both buckets to water the flowers. How many gallons of water does she use?
Answer:
Number of gallons she uses both buckets to water the flowers = 4.95.

Explanation:
Number of gallons one bucket she has = 3.4 gallons.
She has one bucket with 3.4 gallons of water and another with 1.85 gallons less.
=> Number of gallons other bucket she has = Number of gallons one bucket she has – 1.85
=> 3.4 – 1.85
=> 1.55.
Number of gallons she uses both buckets to water the flowers = Number of gallons one bucket she has + Number of gallons other bucket she has
= 3.4 + 1.55
= 4.95.