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Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.2 Reflections to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.2 Answer Key Reflections

Math in Focus Grade 8 Chapter 8 Lesson 8.2 Guided Practice Answer Key

Solve.

Question 1.
Andrew wants to hang a square poster on his bedroom wall. He knows that to get the poster to balance properly, he needs to place a second picture hanger at W’, with the dotted vertical line as the line of symmetry. If the distance between W and the vertical line is 3 inches, find WW’.

Answer:
W’ is the reflection of W so if W is 3 inches from the vertical then W’ is also 3 inches away from the vertical line. Thus WW’is the sum of the length of W and the length of W’. Therefore, WW’ is 6 inches.

Technology Activity

Explore The Properties Of Reflections With Geometry Software

Step 1.
Draw a vertical line segment using a geometry software program.

Step 2.
Select the y-axis as the line of reflection. Reflect the line segment. How are the line and its image related to the line of symmetry?

Step 3.
How are the lengths of the segment and its image relate?

Step 4.
Repeat Step 1 to Step 3 first using a horizontal segment and then a segment that is neither vertical nor horizontal.

Step 5.
Repeat Step 1 to Step 4 using the x-axis as the line of reflection.

Step 6.
Draw a rectangle and reflect it over the y-axis. Do the parallel lines of the rectangle stay parallel? Do the perpendicular sides of the rectangle stay perpendicular?

Step 7.
Repeat Step 6 using the x-axis as the line of reflection.

Math Journal Which of these properties does a reflection seem to preserve: length, shape, parallel lines, or perpendicular lines? Explain.

Draw Images After Reflections.

In the activity, you may have observed the following:

Reflections preserve shape and size. They also preserve parallelism and perpendicularity.
All polygons and combinations of polygons are made up of line segments joined at their endpoints. When you reflect these figures, you reflect all the line segments joined at their endpoints. The images of all these line segments and their endpoints combine to form the whole image.

Copy and complete on graph paper.

Question 2.
Each line segment is reflected in \(\overleftrightarrow{M N}\). On a copy of the diagram, draw each image.

a)

Answer:

b)

Answer:

c)

Answer:

d)

Answer:

Copy and complete on graph paper.

Question 3.
Layla is designing a star-shaped figure for a stencil. She wants the bottom half to be a reflection of the top half. She will reflect it across the x-axis to draw the other half. Complete the design for her.

Answer:
P(-4, 0), Q(0, 4) and R(4, 0) are plotted on the plot. Put additional points M and N on the coordinates (-1, 1) and (1, 1) to make the plotting easier. Since the reflection is in the x-axis then all the coordinates will not change. For P(-4, 0) and R(4, 0), since 0 is 0 units away from x then for image P’ and R’ will be along the x-axis with coordinates (-4, 0) and (-4, 0) respectively. Point Q(0, 4) on the other hand is 4 units above the x-axis so its R’ units below the axis. For M(-1, 1) and N(1, 1)since it is 1 unit above the x-axis then their image will be 1 unit beneath that axis. Thus M’ will be on (-1, 1) and N’ will be on (1, -1).

Complete.

Question 4.
A figure has vertices P (0, 2), Q (-1, 0), R (-2, 1), S (-1, -2), and T (0, -2), is reflected in the y-axis. Draw the figure and its image on the coordinate plane. Use 1 grid square on both axes to represent 1 unit for the interval from -2 to 2.
Answer:

Complete.

Question 5.
Mr. Patterson is building a double bird house, one next to the other. The vertices of the front of one houses have coordinates P (3, 0), Q (5, 3), R (3, 6), and S (1, 3).
The front of the other bird house, P’Q’R’S’, is a reflection of the first one in the y-axis.
The x-coordinates of vertices of PQRS and P’Q’R’S’ are , and their y-coordinates are .
P(3, 0) is mapped onto P'(, ).
Q (5, 3) is mapped onto Q’ (, ).
R (3, 6) is mapped onto R’ (, ).
S (1, 3) is mapped onto S’ (, ).
Any point (x, y) is mapped onto (, ) when reflected in the y-axis.

Answer:

The x-coordinates of vertices of PQRS and P’Q’R’S’ are negative of each other and their y-coordinates are did not change.
P(3, 0) is mapped onto P'(-3, 0).
Q (5, 3) is mapped onto Q’ (-5, 3).
R (3, 6) is mapped onto R’ (-3, 6).
S (1, 3) is mapped onto S’ (-1, 3).
Any point (x, y) is mapped onto (-x, y) when reflected in the y-axis

Math in Focus Course 3B Practice 8.2 Answer Key

Copy each diagram on graph paper and draw the image using the given reflection.

Question 1.
In the x-axis

Answer:

Question 2.
In the y-axis

Answer:

Question 3.
Ethan placed six sticks on a table. Three of the sticks, \(\overline{P Q}\), \(\overline{R S}\), and \(\overline{T U}\) are shown on the coordinate plane. The other sticks are images of the three sticks, with x = 0 as the line of reflection. On a copy of the graph, draw the sticks not shown on the coordinate plane.

Answer:

Solve.

Question 4.
A pattern is drawn on the coordinate plane and then repeated by first reflecting it in the x-axis and reflecting the original pattern in the y-axis.

a) Copy and complete the table by finding the position of each of the other tiles. On a copy of the coordinate plane, indicate the positions of the images.

Answer:
Note that if A, B, C, D, E, and Fis reflected in x-axis, the coordinates wiLl be in the form (x, -y) while the image after applying the reflection in y-axis has coordinates in the form (-x, y) then and in the coordinate plane.

b) Reflect A’, B’, and C’ in the y-axis. What are the coordinates of the image?
Answer:
If A’, B’, and C’ is reflected in y-axis, the coordinates wiLL be in the form (-x, y) then the ¡mage of A’(4, -5) will be on (-4, -5), B’(4, -4) will have an image on (-4, -4) and the image of C’(5, -4) will be plotted on (-5, -4).

c) Reflect A”, B”, and C” in the x-axis. What are the coordinates of the image? How do these coordinates compare to those in b)?
Answer:
If A”,B”, and C” ¡s refLected in x-axis, the coordinates will be in the form (x, -y) then the images of A”(-4, 5), B”(-4, 4), and C”(-5, 4) will be on (-4, -5), (-4, -4) and (-5, -4). Observe that A’, B’, and C’ if reflected in y-axis has the same image with A”, B”, and C” if it is reflected in x-axis.

Copy and complete on graph paper.

Question 5.
Isabella painted a water color design on graph paper. Some of the points were at A (-4, 8), 8 (-2, 8), C(-1, 6), D (-2, 4), E (-4, 4), and F(-5, 6). She folded the paper along y = 3 to reflect the design.
The image points are A’, 8′, C’, D’, E’, and F’.

a) Draw the line y = 3.
Answer:

The line y equals to 3 is a line segment whose coordinates are (0,3) then

b) Find the coordinates of A’, B’, C’, D’, E’, and F’.
Answer:
A coordinates are (-4, 8) first know the position of the point on the coordinate plane. Notice that A is 5 units above y = 3, so the y coordinate for the image A is 5 units below y = 3 which is -2. Therefore, the coordinates of A’ is (-4 ,-2).
B coordinates are (-2, 8) first locate the point on the coordinate plane. Notice that B is 5 units above y = 3, so the y coordinate for the image B is 5 units below y = 3 which is -2. Therefore, the coordinates of A’ is (-2, -2).

Locating C(-1, 6) observe that C is 3 units above y = 3 then, y coordinate for the image C is 3 units under y = 3 which is 0. Therefore, the coordinates of C’ is (-1, 0).

For D whose coordinates are (-2, 4), the first step is to locate the point notice that D is 1 unit above y = 3 so the y coordinate for the image of D is 1 unit below y = 3 which is 2. Therefore, the coordinates of D’ is (-2, 2).

Determine the position of E (-4, 4) note that E is 1 unit above y = 3 thus, the y coordinate for the image of E is 1 unit beneath y = 3 which is 2. Therefore, the coordinates of E’ is (-4, 2).

Lastly, Locate F(-5, 6). Observe that F is 3 units above y = 3 then, the y coordinate for the image of F is 3 units under y = 3 which is 0 . Therefore, the coordinates of F’ is (-5, 0).

c) Draw the image and label A’, B’, C’, D’, E‘, and F’.
Answer:

Solve.

Question 6.
The image of a butterfly with its wings symmetrically spread out is outlined on the coordinate plane. The uppermost tips of the wings are at (4, 5) and (-2, 5). The lower most tip of one wing is at (2, 0).

a) Find an equation of the line of reflection.
Answer: Plot the points on the coordinate plane. Let these points (4, 5) and (-2, 5) be an original point and its image. The y coordinates did not change, the reflection must be in x. Also, they are 6 points away from each other, the points are 3 units away from x = 1. Since the line of reflection is exactly in between the original point and its image, therefore, the line of reflection for this case is x =1.

b) Find the position of the lower tip of the other wing.
Answer: The line of reflection x = 1, let (2, 0) be an original point and the image be the point where the last tip of the wing is. Observe that the line of reflection is a vertical line then the image of (2, 0) will be on the same horizontal line where (2, 0). Next, the original point is 1 unit away from the line of reflection thus its image is 1 unit from the opposite side of x =1. Therefore, on the same horizontal line, move 1 unit to the left the line of reflection, the other tip of the wings is on (0, 0).

Question 7.
Math Journal Point A’ is the image of point A under a reflection. How do you find the line of reflection, without the use of a coordinate grid?
Answer: The reflection of A has the same distance from the line of reflection. Therefore, the line of reflection can be determined by adding the lengths of the original point and the image then dividing the sum by 2. Their midpoint must be the line of reflection.

Question 8.
A tablecloth has two red dots on it. They are at positions (-3, -1) and (-1, -3). The cloth is folded in half, so that the dots touch each other. What is an equation for the line along which the tablecloth was folded?
Answer:
x = y
The point (-2,-2) is the reflection of the points (-3,-1) and (-1,-3) since it is exactly in between the points. Both are 1 diagonal unit away from the point. Thus, the diagonal line whose points are (-3,-3),(-2,-2),(-1,-1)is the line of reflection of the points. When y is 1, x is equal to 1, if y is equal to 2 x is 2.

Question 9.
A leaf is symmetric about its midvein, the central vein that runs the length of the leaf. The leaf is outlined in the coordinate plane with its midvein on the line y = -x.
a) A side vein has a length of 6 units on the grid. What is the length of its symmetric counterpart?
Answer:
6 units
The line of reflection is exactly in the middle of the original point or line and its image. If the original points or lines are 10 units away from the line of reflection, then so is the image, if it is 1 unit away, then so is their reflection. Let the side vein be the original line, the grid be the reflection of the line and the image be the symmetric counterpart of the vein. From the above explanation, since the original line is 6 units away from the grid then the image is also 6 units away.

b) The endpoint of another side vein is at (4, 3). What is the endpoint of its symmetric counterpart?
Answer: (-3,-4)
Explanation:
The line of reflection according to its equation is a diagonal line. Plotting (4,3) notice that the point is 3 and a half diagonal units from the line of reflection thus its image also has the same distance on the opposite side. So from the line of reflection, move 3 and half diagonal units. Therefore, the endpoint of the corresponding symmetric side of the vein is (-3,-4).

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.1 Translations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.1 Answer Key Translations

Math in Focus Grade 8 Chapter 8 Lesson 8.1 Guided Practice Answer Key

Copy and complete on graph paper.

Question 1.
Abigail jogs from point H (3, 2) in a park to point H’, as described by a translation of 5 units to the left and 3 units up. Mark the position of H’ on the coordinate plane.

Answer:

Therefore, H point is on (-2,5)

Copy and complete on graph paper.

Question 2.
Mr. McBride wanted to set up a barbeque pit in his backyard. He had to move a swing set represented by \(\overline{H L} \text {. }\). He decided to move the swing set by a translation of 4 units to the right and 2 units down to \(\overline{H^{\prime} L^{\prime}}\). Draw and mark \(\overline{H^{\prime} L^{\prime}}\) on the coordinate plane.

Answer:

Technology Activity

Explore The Properties Of Translations With Geometry Software

Step 1.
Draw a line segment using a geometry software program.
Step 2.
Select the Translate function, within the Transform menu. Enter the number of units by which you wish to translate the line segment. Translate the line segment to its new position.
Step 3.
How is the length and position of the new line segment related to the original line segment?

Think Math
A horizontal segment is translated horizontally. Is the image parallel to the original segment? Does a translation always map a segment onto a parallel segment? Explain.

Step 4.
Repeat Step 2 using a triangle and then a rectangle. Observe how each figure is related to its image.

Step 5.
How do the size and shape of the figures change under a translation? Do the parallel sides of the rectangle remain parallel? Do the perpendicular sides of the rectangle remain perpendicular?

Math Journal
Which of these properties does a translation seem to preserve: lengths, shapes, parallel lines, or perpendicular lines? Explain.

Draw Images After Translations.

From the activity, you have observed that translations preserve shape, size, parallelism, and perpendicularity.

When a figure is drawn on a coordinate plane, you can easily draw its image. You need only to find the images of the vertices and connect them appropriately.

Complete on graph paper.

Question 3.
Tim is creating a wrapping paper design on his computer. He wants to move figure EFGH by translating it 5 units to the left and 4 units down. E, F, G, and FI have the coordinates (1, 2), (3, 2), (3, 3), and (1, 3). Draw EFGH and E’F’G’H’ on the same coordinate plane. Use 1 grid square on both axes to represent 1 unit for the interval from —4 to 3. Then complete the following statements.
E (1, 2) is mapped onto E’ (, ).
F(3, 2) is mapped onto F’ (, ).
G (3, 3) is mapped onto G’ (, ).
H (1, 3) is mapped onto H’ (, ).
Answer:
E (1, 2) is mapped onto E’ (-4,-2).
F(3, 2) is mapped onto F’ (-2,-2).
G (3, 3) is mapped onto G’ (-2,-1).
H (1, 3) is mapped onto H’ (-4,-1).

Complete.

Question 4.
A triangle has coordinates A (2, 1), B (3, 2), and C (1, 4). It is moved under the translation 2 units to the left and 3 units up. Find the coordinates of the image triangle A’B’C’. Then state the new coordinates for any point (x, y) under this translation.

To find the coordinates of A’, B’, and C’, subtract 2 units from the x-coordinate and add 3 units to the y-coordinate of A, B, and C.

Answer:

Math in Focus Course 3B Practice 8.1 Answer Key

Find the coordinates of the image under each translation.

Question 1.
P (0, 2) is translated by 8 units to the left.
Answer: P is mapped onto P’ (-8,2).

Question 2.
Q(-3, 5) is translated by 3 units to the right and 10 units up.
Answer: Q will be (0, 15) of the coordinate plane.

Question 3.
R(-4, -2) is translated by 1 unit to the left and 6 units up.
Answer: R(-4, -2) is mapped onto R'(-5, 4).

Copy each diagram on graph paper and draw the ¡mage under each translation.

Question 4.
\(\overline{A B}\) is translated 5 units to the right and 1 unit down.

Answer:

Question 5.
Triangle DEF is translated 3 units to the left and 2 units up.

Answer:

Find the coordinates of each point using the given translation. Label the images on a coordinate plane.

Question 6.
Jon’s apartment is located at A (2, 2). He uses the translations described in a) to d) to visit each of his neighbors.

a) From A (2, 2), translate by 3 units to the right, 2 units up to B.
Answer:

b) From B, translate by 2 units to the left, 1 unit up to C.
Answer:

c) From C, translate by 1 unit to the right, 2 units down to D.
Answer:

d) From D, translate by 2 units to the left, 3 units down to E.
Answer:

Solve. Show your work.

Question 7.
The base of a box is at ABCD. It is moved by a translation to a new position A’B’C’D’. The table shows the position to which A was mapped. Find the new position of the other three vertices of the base. Copy and complete the table.

Answer:

Question 8.
A crane moved a cargo pallet from ABCD to other positions on the ship’s deck.
a) Find the coordinates of A’B’C’D’ under a translation that moves each point (p, q) to (p + 4, q + 1). Copy and complete the table. Draw A’B’C’D’ on a graph paper.

Answer:

b) The position of A”B”C”D” is shown on the coordinate plane. State the new coordinates of any point (x, y) under the translation from ABCD to A”B”C”D”.

Answer:
A,B,C and D are on (-2, 1), (0, 3), (2, 1) and (0, -1), A” will have coordinates (-6, 4), and B” has (-4, 6). On the other hand, C” will be lying on (-2, 4) and D” is on (-4, 2).

Question 9.
A computer program T instructs a robotic arm to move an object on the coordinate plane 2 units to the right and 3 units down. The object at point P is translated by T to point P’. Find the coordinates of P if point P’ is (3, 3).
Answer: The original point for P'(3, 3) is P(1, 6).

Question 10.
A line has the equation y = x. It is translated up by 3 units. What is the equation of the new line? How do the slopes of the line and its image compare?
Answer:

Both the original and new line segment are linear but y + 3 = x is 3 units above y = x.

Question 11.
In a wallpaper pattern, a vertical stripe at x = -1 is copied by moving it to x = 1. Describe the translation of this stripe both verbally and algebraically.
Answer:
First locate at x at -1 on the coordinate plane, then know the position of the point at 1. The distance of the two would determine the number of units it move, so the translation is 2 units. Also the original points is x at -1 and then it moved it to right. Therefore, the translation is 2 units to the right.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Geometric Transformations to score better marks in the exam.

Math in Focus Grade 8 Course 3 B Chapter 8 Answer Key Geometric Transformations

Math in Focus Grade 8 Chapter 8 Quick Check Answer Key

Solve. Show your work.

Question 1.
Points W and Z are reflections of each other on the x-axis. If W is the point (4, -3), what are the coordinates of Z?
Answer:
Given that the points W and Z are reflections of each other on the x-axis.
The points of W are (4,-3)
The points (4,-3) are in the 4th quadrant.
The coordinates of Z are (4,3) because the reflection to the fourth quadrant in the x axis is the first quadrant. The points (4,3) lie in the first quadrant.

Question 2.
Points P (-2, 5) and Q (-2, -5) are reflections of each other in the line k. What is an equation of line k?
Answer:
Given that the points are P(-2,5) and Q(-2,-5)
(x1, y1) = (-2,5)
(x2,y2) = (-2,-5)
We know that, Equation of the line formula is y2 – y1/x2 – x1
Equation of the line k = y2 – y1/x2 – x1
k = -5-(5)/-2-(-2)
k = -5-5/-2+2
k = -10/0
k = -10.

Question 3.
Points M (2, -1) and N (-2, -1) are reflections of each other in the line m. What is an equation of line m?
Answer:
Given points are M(2,-1) and N(-2,-1)
Here (x1,y1) = (2,-1)
(x2,y2) = (-2,-1)
We know that, Equation of the line formula is y2 – y1/x2 – x1
equation of the line m = y2 – y1/x2 – x1
m = -1-(-1)/-2-(2)
m = -1+1/-4
m = 0/-4
Therefore m = 0

Question 4.
If a point S is the reflection of R (1,4) in the x-axis, what is the length of \(\overline{R S}\)?
Answer:
Given point is R(1,4)
The point R is reflection to S(1,-4)
The distance between the RS is 8 grids.

State whether x and y are directly proportional.

Question 5.
y = 1.5x
Answer:
Given equation is y = 1.5x
Suppose x,y = 2
Substitute in the above equation
2 = 1.5 × 2
2 = 3
Here x increases then y also increases so, the given equation is directly proportional.

Question 6.
y = –\(\frac{x}{2}\)
Answer:
Given equation is y = -x/2
Suppose x,y = 2
Substitute in the above equation
2 = -2/2
2 = -1
Here x increases then y decreases so, it is not directly proportional.

Question 7.

Answer:
x = y²
We take x = 2 and y = 2
Then 2 = 2²
2 = 4
Here x increases then y decreases so, it is directly proportional.

State whether \(\overleftrightarrow{Y Z}\) is a perpendicular bisector of the given segment. Justify your answer. If \(\overleftrightarrow{Y Z}\) is a perpendicular bisector, name two pairs of distances that are equal.

Question 8.

Answer:
In the given figure CD and ZY are not perpendicular bisectors.
Perpendicular bisector means it is a line or a segment that passes through the midpoint of the segment. And any point on the perpendicular bisector is equidistant from the endpoints of the line segment.
Here the two pairs CD and ZY distances are not equal.

Question 9.

Answer:
In the given figure EF and YZ are perpendicular bisectors.
Perpendicular bisector means it is a line or a segment that passes through the midpoint of the segment. And any point on the perpendicular bisector is equidistant from the endpoints of the line segment.
Here the two pairs EF and YZ distances are equal.

Question 10.

Answer:
In the given figure GH and YZ are perpendicular bisectors.
Perpendicular bisector means it is a line or a segment that passes through the midpoint of the segment. And any point on the perpendicular bisector is equidistant from the endpoints of the line segment.
Here the two pairs GH and YZ distances are equal.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 7 Review Test to score better marks in the exam.

Math in Focus Grade 8 Course 3 B Chapter 7 Review Test Answer Key

Concepts and Skills

For this review, you may use a calculator. Use 3.14 as an approximation for π.
Round your answer to the nearest tenth where necessary.

Question 1.
Find the value of x in each of the following diagrams.
a)

Answer:
The value of x is 17.69 approximately 18 in,

Explanation:
First we find the altitude a for smaller right traiangle,
a² + 6² = 10²,
a² = 100 -36 = 64 so a is square root of 64 is 8 inches as bigger right triangle
altitude is twice the smaller one so it is 2 X 8 =16 inches now
x² = 12² + 16²,
x² = 144 + 169 = 313, therefore x is square root of 17.69 approximately 18 in.

b)

Answer:
The value of x is 12 cm,

Explanation:
Given two right triangles first we calculate hypotenuse be y of
right triangle of width 12.8 cm and height  9.6 cm so
y ² = 9.6² + 12.8² = 92.16 + 163.84 = 256,
so the value of y is square root of 256 is 16 cm,
Now the value of x is  x² + 16² = 20²,
x² = 400 – 256 = 144 , the value of x is square root of 144 is 12 cm.

c)

Answer:
The value of x is approximately equal to 19 cm,

Explanation:
Applying pythagorean we have
x² + 7² = 20²,
x² = 400 – 49 = 351, the value of x is square root of 351 is 18.734,
so x is approximately equal to 19 cm.

Question 2.
Find the distance between each pair of points. Which pair of points are the greatest distance apart?
a) A(5, 2), B(8, 5)
Answer:
To find the distance of two points, use the distance formula given by the square root of the sum of the x-coordinate of the second point minus the x-coordinate of the first point and the y-coordinate of the second point minus the y-coordinate of the first point Also. note that there is no negative magnitude(distance) Then, let A(5, 2) be (x₁, y₁) whiLe B(8, 2) be (x₂, y₂) and e be the distance of A to B.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(8-5)^{2}+(2-2)^{2}}\)    Substitution.
= \(\sqrt{3^{2}+0^{2}}\)    Subtract.
= \(\sqrt{9+0}\)    Evaluate the squares.
= \(\sqrt{9}\)    Add.
= ±3 Evaluate.
Therefore, A is 3 units away from B.

b) C(-3, 2), D(2, 3)
Answer:
Apply the distance formula then Let C(-3, 2) be (x₁, y₁) while D(2, 3) be (x₂, y₂) and f be the distance of C to D.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(2-(-3))^{2}+(3-2)^{2}}\)    Substitution.
= \(\sqrt{5^{2}+1^{2}}\)    Subtract.
= \(\sqrt{25+1}\)    Evaluate the squares.
= \(\sqrt{26}\)    Add.
≈ ±5.1    Evaluate.
Therefore, the distance of C to D is 5.1 units

c) L(1, -3), N(-2, -1)
Answer:
Use the distance formula then let L(1, -3) be (x₁, y₁) white N(-2, -1) be (x₂, y₂) and g be the distance of L to N.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{((-2)-1)^{2}+((-1)-(-3))^{2}}\)    Substitution.
= \(\sqrt{(-3)^{2}+2^{2}}\)    Subtract.
= \(\sqrt{9+4}\)    Evaluate the squares.
= \(\sqrt{13}\)    Add.
≈ ± 3.6     Evaluate.
Therefore, L is 3.6 units away from N.

d) Y(-3, -3), Z(0, 4)
Answer:
Let Y(-3, -3) be (x₁, y₁) while Z(0, 4) be (x₂, y₂) and h be the distance of Y to Z then apply the distance formula.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(0-(-3))^{2}+(4-(-3))^{2}}\)    Substitution.
= \(\sqrt{3^{2}+7^{2}}\)    Subtract.
= \(\sqrt{9+49}\)    Evaluate the squares.
= \(\sqrt{58}\)    Add.
≈ ± 7.6     Evaluate.
Therefore, Y is 7.6 units away from Z.

Since A is 3 units away from B. the distance of C to D is 5.1 units, L is 3.6 units away from N. and Y is 7.6 units away from Z. Therefore, the points with largest distance is Y and Z.

Question 3.
Find the volume of each of the following composite solids.
a)

Answer:
The volume of given solid is 120 in³,

Explanation:
First we calculate x,
x² = 6² + 8²,
x² = 36 + 64 = 100, the value of x is square root of 100 is 10 in,
Now the volume of given soild is \(\frac{1}{3}\) • (base edge)² X h =
\(\frac{1}{3}\) • (6)² X 10 in = 120 in³.

b)

Answer:
The volume is 3704.094 in³,

Explanation:
First we calculate value of  x which is height of cylinder,
Applying pythagorean theorem 16² = 9.6² + x²,
x² = 256 – 92.16 = 163.84, the value of x is square root of
163.84 is 12.8 in now volume of cylinder is πr²h =
3.14 X 9.6 X 9.6 X 12.8 = 3704.094 in³.

c)

Answer:
The volume is 113.04 in³,

Explanation:
We have height of sphere is 2r, so r = 6/2 = 3 in,
Now volume of sphere is \(\frac{4}{3}\) • πr³ = \(\frac{4}{3}\) X 3.14 X 3 X 3 X 3 = 113.04 in³.

Question 4.
Find the value of x in each of the following diagrams.
a)

Answer:
To solve for the voLume of the composite soLid, add the volume of the cyLinder to the voLume of the cone. For the volume of the cylinder,
Volume of Cylinder = Area of the base · height of Cylinder
= 20 · 0.5
= 10
Thus, the cylinder has a volume of 10 cubic centimeters. Next, solve for the volume of the cone. First find the radius of its base. Notice that the radius r of the base, the slanted side s and the height j of the cone forms a right triangle with the slanted side as the hypotenuse. Use the Pythagorean Theorem which describes the sum of the square of the legs of a right triangle to be equal to the square of the hypotenuse.
j² + r² = s²      Use the Pythagorean Theorem.
(2.8)² + r² = (3.2)²      Substitution
7.84 + r² = 10.24    Multiply.
7.84 + r² – 7.84 = 10.24 – 7.84     Subtract 7.84 to both sides.
r² = 2.4    Simplify.
r = \(\sqrt{2.4}\)    Find the positive square root.
r ≈ 1.5     Simplify.
Thus, radius r measures around 1.5 cm. Now solving for the volume of the cone, note that the area of a circle (which is the base of the cone) is pi multiplied to the square of the radius and use 3.14 as the value of π.
Volume of Cone = \(\frac{1}{3}\) · Area of the base · Height of Cone
= \(\frac{1}{3}\) πr² · j
= \(\frac{1}{3}\) (3.14) · (1.5)² · 2.8
≈ 6.6
So, the volume of the cone is about 6.6 cubic centimeters. Now for the total volume of the composite solid,
Volume of the Composite Solid = Volume of the Cone + Volume of the Cylinder
= 6.6 + 10
= 16.6
Therefore, the composite solid has a 16.6 cm³ volume.

b)

Answer:
To solve for the volume of the composite solid, add the volume of the cube to the volume of the triangular prism. For the volume of the cube,
let s be the measure of the sides.
Volume of Cube = s³  Formula.
= 6³     Substitution.
= 216    Evaluate.
Thus, the volume of the cube is 216 cubic centimeters. Next for the triangular prism, first solve for the measure of its base. Notice that the edges of the triangular prism has a 1.5 centimeters distance from the ends of the side of the cube. Then the length of the side of the triangular base of the prism can be computed.
Measure of the Base = Measure of the Side of the Cube + 1.5 + 1.5
= 6 + 1.5 + 1.5
= 9
So, the side of the base of the triangular prism measures 9cm. Then, find the altitude of the triangular base of the prism. Notice that half of the length of the base side of the prism denoted as e, the slanted side s, and the altitude a creates a right triangle with the slanted side as the hypotenuse. Then, applying Pythagorean Theorem states that the sum of the square of the legs of a right triangle is equal to the square of the hypotenuse,
(\(\frac{e}{2}\))² + a² = s²    Use the Pythagorean Theorem.
(\(\frac{9}{2}\)) + a² = (7.5)²     Substitution.
(4.5)² + a² = (7.5)²    Divide.
20.25 + a² = 56.25    Multiply.
20.25 + a² – 20.25 = 56.25 – 20.25   Subtract 20.25 to both sides.
a² = 36    Evaluate.
a = \(\sqrt{36}\)    Find the positive square root.
a = 6 Simplify.
Thus, the altitude of the triangLe is 6 centimeters Then, note that the area of a triangle (which is the base of this prism) is half of its length multiplied by the aLtitude. Also, notice that the side of the cube is equal to the height of the triangular prism so the height h of the prism is equal to 6 centimeters. Solving for the volume of the triangular prism,
Volume of Triangular Prism = Area of the Base · Height of the Prism
= \(\frac{1}{2}\) · e a · h
= (\(\frac{1}{2}\) · 9 · 6) · 6
= 27 · 6
= 162
Thus, the triangular prism has a volume of 162 cubic centimeters. Now, for the total volume of the composite solid,
Volume of the Composite Solid = Volume of the Cube + Volume of the Triangular Prism
= 216 + 162
= 378
Therefore, tne composite solid has a 378 cm³ volume.

Problem Solving

Solve. Show your work.

An interactive game teaches users how to find the distance between two points on a coordinate plane. Each unit on the grid equals 1 kilometer. Use the grid for questions 5 and 6.

Question 5.
Find the distance between the farmer and each location.
a) River
b) Pond
c) Town
d) Orchard

Answer:
a) To find the distance of two points, use the distance formula given by the square root of the sum of the x-coordinate of the second point minus the x-coordinate of the first point and the y-coordinate of the second point minus the y-coordinate of the first point Also, note that there is no negative magnitude(distance). Then, let Mr. Farmer at (1, 2) be (x₁, y₁) white the river at (5,4) be (x₂, y₂) and e be the distance of Mr Farmer and the river.
e = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(5-1)^{2}+(4-2)^{2}}\)     Substitution.
= \(\sqrt{4^{2}+2^{2}}\)      Subtract.
= \(\sqrt{16+4}\)     Evaluate the squares.
= \(\sqrt{20}\)     Add.
≈ ± 4.5 Evaluate.
Therefore, Mr. Farmer is approximate[y 4.5 kilometers away from the river.

b. Apply the distance formula again letting Mr. Farmer at (1, 2) be (x₁, y₁) and pond (-4, -2) be (x₂, y₂) and f be the distance of the pond and Mr. Farmer
f = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{((-4)-1)^{2}+((-2)-2)^{2}}\)     Substitution.
= \(\sqrt{(-5)^{2}+(-4)^{2}}\)      Subtract.
= \(\sqrt{25+16}\)     Evaluate the squares.
= \(\sqrt{41}\)     Add.
≈ ± 6.4   Evaluate.
Therefore, the distance of Mr. Farmer to the pond is about 6.4 kilometers.

c) Use the distance formula then let Mr. Farmer at (1, 2) be (x₁, y₁) while the town at (-4, 5) be (x₂, y₂) and g be the distance of Mr. Framer to the town.
g = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{((-4)-1)^{2}+(5-2)^{2}}\)     Substitution.
= \(\sqrt{(-5)^{2}+3^{2}}\)      Subtract.
= \(\sqrt{25+9}\)     Evaluate the squares.
= \(\sqrt{34}\)     Add.
≈ ± 5.8     Evaluate.
Therefore, Mr Farmer is around 5.8 kilometers away from the town

d) Let Mr Farmer (1, 2) be (x₁, y₁) while Orchard (6, -4) be (x₂, y₂) and h be the distance of Mr. Farmer and the orchard then apply the distance formula.
h = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)    Formula.
= \(\sqrt{(6-1)^{2}+((-4)-2)^{2}}\)     Substitution.
= \(\sqrt{5^{2}+(-6)^{2}}\)      Subtract.
= \(\sqrt{25+36}\)     Evaluate the squares.
= \(\sqrt{61}\)     Add.
≈ ± 7.8     Evaluate.
Therefore, Mr Farmer is approximately 7.8 kilometers away from the orchard.

Question 6.
The farmer wants to go fishing in either the river or the pond. Which one is closer to his current position? How many kilometers fewer will he walk if he chooses the source that is closer?
Answer:
The objective is to determine the which body of water is closer to Mr. Farmer at (1, 2) and the difference of the distance of the bodies of water to one another.

First, determine the distance of the farmer to the first body of water which is the river. Apply the distance formula given by the square root of the sum of the x-coordinate of the second point minus the x-coordinate of the first point and the y-coordinate of the second point minus the y-coordinate of the first point. Also, note that there is no negative magnitude(distance). Then, Let Mr. Farmer at (1, 2) be (x₁, y₁) while the river at (5, 4) be (x₂, y₂) and p be the distance of Mr. Farmer and the river
p = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\) Formula.
= \(\sqrt{(5-1)^{2}+(4-2)^{2}}\)    Substitution.
= \(\sqrt{4^{2}+2^{2}}\)      Subtract.
= \(\sqrt{16+4}\)      Evaluate the squares.
= \(\sqrt{20}\)    Add.
≈ ± 4.5 Evaluate.
Thus, Mr. Farmer is about 4.5 units away from the river

Next, determine the distance of Mr. Farmer to the pond by using the distance formula again Letting Mr. Farmer at (1, 2) be (x₁, y₁) pond (-4, -2) be (x₂, y₂) and y be the distance of the pond and Mr. Farmer
y = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\) Formula.
= \(\sqrt{((-4)-1)^{2}+((-2)-2)^{2}}\)    Substitution.
= \(\sqrt{(-5)^{2}+(-4)^{2}}\)      Subtract.
= \(\sqrt{25+16}\)      Evaluate the squares.
= \(\sqrt{41}\)    Add.
≈ ± 6.4    Evaluate.
Thus, the distance of Mr. Farmer to the pond is around 6.4 units. Since the distance of the river to Mr. Farmer is smaller than the distance of the pond to Mr Farmer, therefore, the river is cLoser to his position.

Lastly, to determine the difference of the distance, subtract the distance of the farmer to the river to the distance of the farmer to pond.
Distance to the Pond – Distance to the River
6.4 – 4.5
1.9
Therefore, the difference of the distance of the pond and river to Mr. Farmer is 1.9 units.

Question 7.
Candy builds a cage for some chickens out of wood and wire mesh, as shown below. The two slanted sides of the cage are the same size and shape. Find the height of the cage.

Answer:
The height of the cage is 24 in,

Explanation:
Given Candy builds a cage for some chickens out of wood and
wire mesh, as shown the two slanted sides of the cage are
the same size and shape, we take width as 64/2= 32 in and consider
right triangle and apply pythagorean theorem to find height h of the cage as
h² + 32² = 40²,
h² = 1600 – 1024 = 576, so h is square root of 576 is 24 in.

Question 8.
A straw that is ‘12 centimeters long fits inside a glass. The height of the glass is 7.5 centimeters. Find the radius of the glass.

Answer:
The radius of the glass is 4.68 centimeters,

Explanation:
Given a straw that is ‘12 centimeters long fits inside a glass.
The height of the glass is 7.5 centimeters, let b be the base of the glass,
applying pythagorean theorem
b² + 7.5² = 12²,
b² = 144 – 56.25 = 87.75,therefore b is square root of 87.75 is 9.36 centimeters,
Now the radius of glass is half base is 9.36/2  = 4.68 centimeters.

Question 9.
Naomi makes a lampshade by making a cone and then cutting off
part of the cone, as shown. Find the diameter of the larger opening in the lampshade.

Answer:
The diameter of the larger opening in the lampshade is 5.71 in,

Explanation:
Applying pythagorean theorem let d be diameter of the
larger opening in the lampshade so d² + 20² = 20.8²,
d² = 432.64 – 400 = 32.64 ,so d is square root of 32.64 is 5.71 in.

Question 10.
Four identical pieces are cut off the sides of a cylinder, as shown. The remaining shape is a square prism. The diagonal of the prism’s square base is as long as the diameter of the cylinder. The diameter of the cylinder is 12 centimeters, and the height of the cylinder is 9 centimeters. Find the length of a side of the square base of the prism. Then find an approximate volume of the prism.

Answer:
The length of a side of the square base of the prism 3.82 cm,
Approximate volume of the prism is 67 cm³,

Explanation:
Given Four identical pieces are cut off the sides of a cylinder, as shown.
The remaining shape is a square prism. The diagonal of the prism’s
square base is as long as the diameter of the cylinder.
The diameter of the cylinder is 12 centimeters, and
the height of the cylinder is 9 centimeters.
we have Diameter of cylinder = diagonal of base of square prism
=√2 a, so a = 12/3.14 = 3.82 cm,,
now volume of prism is v = a²h = 3.82 X 3.82 X 4.6 = 67.125 cm³.

Question 11.
Ellen’s computer screen is a 20 inch screen, which means that the length of the diagonal of the rectangular screen is 20 inches. The screen can be laid flat in a box that is 12 inches wide. How long is the box?
Answer:
The box is 16 inch long,

Explanation:
Given Ellen’s computer screen is a 20 inch screen,
which means that the length of the diagonal of the rectangular screen is 20 inches.
The screen can be laid flat in a box that is 12 inches wide. Applying
pythagorean theorem let l is the length of the box so
l² + 12² = 20²,
l² = 400 -144 = 256, so length is square root of 256 is 16 inch.

Question 12.
An 18-centimeter tall suitcase is shaped like a rectangular prism. The length of the diagonal of one of its largest faces is 30 centimeters. Find the width of the suitcase.
Answer:
The width of the suitcase is 24 centimeter,

Explanation:
Given an 18-centimeter tall suitcase is shaped like a rectangular prism.
The length of the diagonal of one of its largest faces is 30 centimeters.
Let w be the width of the suitcase applying pythagores theorem
w² + 18² = 30²,
w² = 900 – 324 = 576, so width is square root of 576 is 24 centimeter.

Question 13.
Vera kicks a ball 123 feet diagonally across a rectangular playground of width 85 feet. Find the length of the playground.
Answer:
The length of the playground is 88.9 feet,

Explanation:
Given Vera kicks a ball 123 feet diagonally across a
rectangular playground of width 85 feet.
Let l be the length of the playground applying pythagorean theorem
l² + 85² = 123²,
l² = 15129 – 7225 = 7904, the value of l is square root of 7904 is 88.9 feet.

Question 14.
A river is 10.5 meters wide and its banks are parallel to each other. John tries to swim straight across, but the current pushes him downstream so that he lands 14 meters from the spot he wanted to reach. How far did he swim?
Answer:
Far did Jhon swim is 17.5 meters,

Explanation:
Given a river is 10.5 meters wide and its banks are parallel to each other.
John tries to swim straight across, but the current pushes him downstream so that he lands 14 meters from the spot he wanted to reach. Applying pythagorean theorem to find how far did he swim s is
s² = 14² + 10.5² = 196 + 110.25=306.25, the value is s square root of 306.25 is 17.5 meters.

Question 15.
An extendable ladder 9.8 feet long leans against a wall with the base 3 feet from the wall. Assuming the base of the ladder does not move, about how much higher will the top of the ladder be above the ground when the ladder extended to twice its original length?
Answer:
higher will the top of the ladder above the ground is 10.04 feet,

Explanation:
Given an extendable ladder 9.8 feet long leans against a wall with the base 3 feet from the wall and assuming the base of the ladder does not move, about how much higher will the top of the ladder be above the ground when the ladder extended to twice its original length, Let h1 is 9.8 feet and h2 is extended length is 2 X 9.8 = 19.6, Applying pythagorean theorem
h₁² =(9.8) ² -(3)²  ( For right triangle)  = 96.04 – 9 = 87.04, so h1 is square root
of 87.04 is 9.32 feet
and h₂² = (19.6)² – (3)² = 384.16 – 9 = 375.16, So h2 is the square root of
375.16 is 19.36 feet, therefore higher will the top of the ladder above
the ground is 19.36 feet – 9.32 feet = 10.04 feet.

Question 16.
A doorway is 80 inches tall and 32 inches wide. Can a round tabletop with a diameter of 90 inches fit through the doorway? If not, what is the greatest possible diameter that will fit through the doorway? Explain.
Answer:
The round tabletop does not fit through the doorway even diagonally,

Explanation:
Given a doorway is 80 inches tall and 32 inches wide,
the diagonal measure d² = 80² + 32² = 6400 + 1024 = 7424,
So d is squareroot of 7424 is 86.16 inches,
So, the round tabletop does not fit through the doorway even diagonally,
The greatest possible diameter of the round doorway is about 86 inches.

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 5 Practice 5 Probability as a Fraction to finish your assignments.

Math in Focus Grade 4 Chapter 5 Practice 5 Answer Key Probability as a Fraction

Find the probability as a fraction in simplest form.

Jake spins the spinner once. He wants to land on these numbers. What is the probability of a favorable outcome? Find the probability as a fraction in simplest form for each outcome.

Example
He wants to land on a number less than 3.
There are 2 favorable outcomes: 1 and 2
There are 8’ possible outcomes: 1, 2, 3, 4, 5, 6, 7, and 8

Question 1.
He wants to land on the number 7.
Answer:
The probability of a favorable outcome is \(\frac{7}{8}\).

Explanation:
Here, the possible outcomes are \(\frac{1}{8}\),\(\frac{2}{8}\),\(\frac{3}{8}\),\(\frac{4}{8}\),\(\frac{5}{8}\),\(\frac{6}{8}\),\(\frac{7}{8}\),\(\frac{8}{8}\). As he wants to land on the number 7 it will be \(\frac{7}{8}\).

Question 2.
He wants to land on an odd number.
Answer:
The possible outcomes are \(\frac{1}{8}\),\(\frac{3}{8}\),\(\frac{5}{8}\),\(\frac{7}{8}\).

Explanation:
Here, the possible outcomes are \(\frac{1}{8}\),\(\frac{2}{8}\),\(\frac{3}{8}\),\(\frac{4}{8}\),\(\frac{5}{8}\),\(\frac{6}{8}\),\(\frac{7}{8}\),\(\frac{8}{8}\). As he wants to land on an odd numbers so the outcomes will be \(\frac{1}{8}\),\(\frac{3}{8}\),\(\frac{5}{8}\),\(\frac{7}{8}\).

Find the probability as a fraction in simplest form for each outcome.

A coin is tossed once. The probability of getting

Question 3.
heads is

Answer:
\(\frac{1}{2}\).

Explanation:
The probability of getting heads is \(\frac{1}{2}\).

Question 4.
tails is
Answer:
\(\frac{1}{2}\).

Explanation:
The probability of getting tails is \(\frac{1}{2}\).

A number cube numbered 1 to 6 is tossed once. The probability of getting

Question 5.
the number 2 is
Answer:
The probability is \(\frac{2}{6}\).

Explanation:
As the number was tossed once, so the probability of getting the number 2 is \(\frac{2}{6}\).

Question 6.
the number 0 is
Answer:
No possible outcomes.

Explanation:
There will be no possibility of getting the number 0 as the number cube numbered 1 to 6.

Question 7.
an even number is
Answer:
The possibilities of getting even number is \(\frac{2}{6}\),\(\frac{4}{6}\),\(\frac{6}{6}\).

Explanation:
The possible outcomes are \(\frac{1}{6}\),\(\frac{2}{6}\),\(\frac{3}{6}\),\(\frac{4}{6}\),\(\frac{5}{6}\),\(\frac{6}{6}\). So the possibilities of getting even number is \(\frac{2}{6}\),\(\frac{4}{6}\),\(\frac{6}{6}\).

Question 8.
a number greater than 4 is
Answer:
The possibilities of getting a number greater than 4 is \(\frac{5}{6}\),\(\frac{6}{6}\).

Explanation:
The possible outcomes are \(\frac{1}{6}\),\(\frac{2}{6}\),\(\frac{3}{6}\),\(\frac{4}{6}\),\(\frac{5}{6}\),\(\frac{6}{6}\). So the possibilities of getting a number greater than 4 is \(\frac{5}{6}\),\(\frac{6}{6}\).

A circular spinner has 4 equal parts. The parts are colored red, blue, green, and yellow. The spinner is spun once. The probability of landing on

Question 9.
red is
Answer:
The probability of landing on red is \(\frac{r}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on red is \(\frac{r}{4}\).

Question 10.
blue is
Answer:
The probability of landing on blue is \(\frac{b}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on blue is \(\frac{b}{4}\).

Question 11.
purple is
Answer:
The probability of landing on purple is \(\frac{p}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on blue is \(\frac{p}{4}\).

Question 12.
green, red, or yellow is
Answer:
The probability of landing on purple is \(\frac{p}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on blue is \(\frac{p}{4}\).

Question 13.
red, blue, green, or yellow is
Answer:
The probability of landing on red, blue, green, or yellow is \(\frac{r}{4}\), \(\frac{b}{4}\), \(\frac{g}{4}\), \(\frac{y}{4}\).

Explanation:
Given that a circular spinner has 4 equal parts and the parts are colored red, blue, green, and yellow and the spinner is psun once. So the probability of landing on red, blue, green, or yellow is \(\frac{r}{4}\), \(\frac{b}{4}\), \(\frac{g}{4}\), \(\frac{y}{4}\).

Find the probability as a fraction in simplest form for each outcome.

A bag contains 10 discs numbered 1 to 10. A disc is drawn from the bag. The probability of drawing

Question 14.
the number 10 is
Answer:
The probability of drawing the number 10 is \(\frac{10}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10. So the probability of drawing the number 10 is \(\frac{10}{10}\).

Question 15.
a number less than 5 is
Answer:
The probability of drawing the number 5 is \(\frac{5}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10. So the probability of drawing the number 5 is \(\frac{5}{10}\).

Question 16.
on odd number is .
Answer:
The probability of drawing the odd number is \(\frac{1}{10}\), \(\frac{3}{10}\), \(\frac{5}{10}\), \(\frac{7}{10}\), \(\frac{9}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So the probability of drawing the odd number is \(\frac{1}{10}\),\(\frac{3}{10}\), \(\frac{5}{10}\), \(\frac{7}{10}\), \(\frac{9}{10}\).

Question 17.
a number divisible by 3 is
Answer:
The probability of drawing the number divisible by 3 is \(\frac{3}{10}\), \(\frac{6}{10}\), \(\frac{9}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So the probability of drawing the number divisible by 3 is \(\frac{3}{10}\), \(\frac{6}{10}\), \(\frac{9}{10}\).

Question 18.
a number greater than 8 is
Answer:
The probability of drawing the number greater than 8 is \(\frac{9}{10}\), \(\frac{10}{10}\).

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So the probability of drawing the number greater than 8 is \(\frac{9}{10}\), \(\frac{10}{10}\).

Question 19.
the number 12 is .
Answer:
There is no possibility of drawing number 12

Explanation:
Here, a bag contains 10 discs numbered 1 to 10 the probability outcomes are \(\frac{1}{10}\), \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), \(\frac{5}{10}\), \(\frac{6}{10}\), \(\frac{7}{10}\), \(\frac{8}{10}\), \(\frac{9}{10}\), \(\frac{10}{10}\). So there is no possibility of drawing number 12.

A bag contains 3 white marbles, 3 blue marbles, and 6 red marbles. A marble is drawn from the bag. The probability of getting

Question 20.
a white marble is
Answer:
The probability of getting a white marble is \(\frac{3}{12}\).

Explanation:
Given that a bag contains 3 white marbles, 3 blue marbles, and 6 red marbles. So the total number of marbles are 3+3+6 which is 12 marbles. So the probability of getting a white marble is \(\frac{3}{12}\).

Question 21.
a blue marble is
Answer:
The probability of getting a blue marble is \(\frac{3}{12}\).

Explanation:
Given that a bag contains 3 white marbles, 3 blue marbles, and 6 red marbles. So the total number of marbles are 3+3+6 which is 12 marbles. So the probability of getting a white marble is \(\frac{3}{12}\).

Question 22.
Which is more likely: drawing a red marble or drawing a blue marble? Explain.
Answer:
Red marbles.

Explanation:
As there are 3 white marbles, 3 blue marbles, and 6 red marbles. So the more likely drawing marble will be red marbles.

Find the probability of each outcome on the number line. Then describe the outcome as more likely, less likely, certain, impossible, or equally likely.

Example
A box contains 4 red pencils, 1 blue pencil, and 1 black pencil. Find the probability of picking a red pencil.

The closer the probability of an outcome is to 1, the more likely the outcome is to occur.

The probability of picking a red pencil is \(\frac{4}{6}\) or \(\frac{2}{3}\)
\(\frac{2}{3}\) is closer to 1 than to 0 on the number line, So, the likelihood of picking a red pencil is more likely.

Each card in a set of 8 cards has a picture of a fruit. There are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. The cards are shuffled, placed in a stack, and one card is picked.

Question 23.
An orange card: ____
Answer:
The probability of picking the orange card is \(\frac{3}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking the orange card is \(\frac{3}{8}\).

Question 24.
An apple card: ____
Answer:
The probability of picking the apple card is \(\frac{2}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking the apple card is \(\frac{2}{8}\).

Question 25.
An apple, peach, or pear card: ____
Answer:
The probability of picking an apple, peach, or pear card \(\frac{2}{8}\), \(\frac{1}{8}\), \(\frac{2}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking an apple, peach, or pear card \(\frac{2}{8}\), \(\frac{1}{8}\), \(\frac{2}{8}\).

Question 26.
An apple, orange, peach, or pear card: ____
Answer:
The probability of picking an apple, peach, or pear card\(\frac{2}{8}\), \(\frac{3}{8}\), \(\frac{1}{8}\) and \(\frac{2}{8}\).

Explanation:
Given that each card in a set of 8 cards has a picture of a fruit and there are 3 orange cards, 2 apple cards, 2 pear cards, and 1 peach card. So the probability of picking an apple, orange, peach, or pear card \(\frac{2}{8}\), \(\frac{3}{8}\), \(\frac{1}{8}\) and \(\frac{2}{8}\).

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 5 Data and Probability to finish your assignments.

Math in Focus Grade 4 Chapter 5 Answer Key Data and Probability

Math Journal

Write the steps to solve the problem.

Neil bought 5 books. The average price of 2 of the books is $5. The average price of the rest of the books is $4. Find the total amount of money Neil paid for the 5 books.

Then, following your steps above, solve the problem.

Answer:
The total amount of money Neil paid for the 5 books is $16.

Explanation:
Given that Neil bought 5 books and the average price of 2 of the books is $5 and the average price of the rest of the books is $4. So the total amount of money Neil paid for the 5 books is
as the price of 2 books is $5 and the price of 1 book is $4. So 5-2 is 3, and the price of the 4 books will be 4×3 which is $12. So the cost for 5 books it will be $12+$4 which is $16.

Put On Your Thinking Cap!

Challenging Practice

Question 1.
Michelle got an average score of 80 on two tests. What score must she get on the third test so that her average score for the three tests is the same as the average score for the first two tests?
Answer:
The score she got on the third test is 80.

Explanation:
Given that Michelle got an average score of 80 on two tests and the sum of score in 2 tests will be 80×2 which is 160. Let the score for third test be x,
so the new average will be \(\frac{160+x}{3}\),
and the average score is 80, so \(\frac{160+x}{3}\) = 80
160+x = 80×3
160+x = 240
x = 240 – 160
= 80.

Question 2.

The line plot shows the shoe sizes of students in Ms. George’s class.

a. How many students are in the class?
Answer:
25 students.

Explanation:
The total number of students are in the class is 25 students.

b. What is the mode of the set of data?
Answer:
3\(\frac{1}{2}\).

Explanation:
The mode of the set of data is 3\(\frac{1}{2}\) as the number that appears most often.

c. How many students in the class wear a size 3\(\frac{1}{2}\) shoe?
Answer:
10 students.

Explanation:
The number of students in the class wear a size 3\(\frac{1}{2}\) shoe is 10 students.

d. Suppose you looked at 100 pairs of shoes for the grade, which includes 3 other classes. How many pairs of size 3\(\frac{1}{2}\) would there be? Explain your answer.
Answer:

Put On Your Thinking Cap!

Problem Solving

Question 1.
The average height of Andy, Chen, and Chelsea is 145 centimeters. Andy and Chen are of the same height and Chelsea is 15 centimeters taller than Andy. Find Andy’s height and Chelsea’s height.
Answer:
The Andy’s height and Chelsea’s height is 140 cm.

Explanation:
Given that the average height of Andy, Chen, and Chelsea is 145 centimeters and Andy and Chen are of the same height and Chelsea is 15 centimeters taller than Andy, so let the height of Andy and Chen be x and the height of Chelsea is 15 centimeters taller than Andy which is x+15. So Andy’s height and Chelsea’s height will be
\(\frac{x+x+x+15}{3}\) = 145
\(\frac{3x+15}{3}\) = 145
3x+15 = 145×3
3x+15 = 435
3x = 435-15
3x = 420
x = 420÷3
= 140.
So the Andy’s height and Chelsea’s height is 140 cm.

Question 2.
Eduardo has 3 times as many stamps as Sally. The average number of stamps they have is 450. How many more stamps does Eduardo have than Sally?
Answer:
\(\frac{1}{2}\) of total number of stamps extra.

Explanation:
Given that Eduardo has 3 times as many stamps as Sally and the average number of stamps they have is 450. Here, if Sally has 1 stamp then Eduardo has 3 stamps. So total stamps will be 4, Eduardo has 2 extra and Eduardo has \(\frac{2}{4}\) which is \(\frac{1}{2}\) of total number of stamps extra.

Question 3.
Bag A and Bag B each contain 2 marbles — 1 white and 1 red. Troy picks 1 marble from Bag A and 1 from Bag B. What is the probability that the following are picked?

a. 2 white marbles
Answer:
\(\frac{2}{4}\)

Explanation:
The probability of picking up 2 white marbles is \(\frac{2}{4}\).

b. 1 red and 1 white marble
Answer:

Explanation:
The probability of picking up 1 red and 1 white marble is

Practice the problems of Math in Focus Grade 4 Workbook Answer Key Chapter 6 Practice 3 Mixed Numbers to score better marks in the exam.

Math in Focus Grade 4 Chapter 6 Practice 3 Answer Key Mixed Numbers

Write a mixed number lor each model.

Example
Write a mixed number for each model

Question 1.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{3}{4}\) = 3 + \(\frac{3}{4}\)
= 3\(\frac{3}{4}\)

Question 2.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{2}{5}\) = 3 + \(\frac{2}{5}\)
= 3\(\frac{2}{5}\)

Write a mixed number for each model.
Question 3.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{1}{2}\) = 3 + \(\frac{1}{2}\)
= 3\(\frac{1}{2}\)

Question 4.

Answer:

Explanation:
1 + \(\frac{3}{5}\) = 1 + \(\frac{3}{5}\)
= 1\(\frac{3}{5}\)

Question 5.

Answer:

Explanation:
1 + 1 + 1 + \(\frac{5}{9}\) = 3 +\(\frac{5}{9}\)
= 3\(\frac{5}{9}\)

Check (✓) the correct model.
Question 6.
Which model shows 1\(\frac{3}{4}\) shaded?

Answer:

Explanation:
Model 1:
1 + 1 + 1 + \(\frac{1}{4}\) = 3 + \(\frac{1}{4}\)
= 3\(\frac{1}{4}\)

Model 2:
1 +  \(\frac{1}{4}\) = 1 +\(\frac{1}{4}\)
= 1\(\frac{1}{4}\)

Question 7.
Which model shows 2\(\frac{3}{5}\) shaded?

Answer:

Explanation:
Model 1:
1 + 1 + \(\frac{3}{5}\) = 2 + \(\frac{3}{5}\)
= 2\(\frac{3}{5}\)
Model 2:
1 + 1 + \(\frac{2}{5}\)  = 2 + \(\frac{2}{5}\)
= 2\(\frac{2}{5}\)

Write each answer as a mixed number.
Question 8.
4 + \(\frac{1}{4}\) = ____
Answer:
4 + \(\frac{1}{4}\) = 4\(\frac{1}{4}\)

4 + \(\frac{1}{4}\)
=4\(\frac{1}{4}\)

Question 9.
3 + \(\frac{5}{9}\) = ____
Answer:
3 + \(\frac{5}{9}\) = 3\(\frac{5}{9}\)

Explanation:
3 + \(\frac{5}{9}\)
= 3\(\frac{5}{9}\)

Question 10.
\(\frac{5}{8}\) + 2 = ____
Answer:
\(\frac{5}{8}\) + 2 = 2\(\frac{5}{8}\)

Explanation:
\(\frac{5}{8}\) + 2
= 2\(\frac{5}{8}\)

Question 11.
\(\frac{3}{5}\) + 4 = ____
Answer:
\(\frac{3}{5}\) + 4 = 4\(\frac{3}{5}\)

Explanation:
\(\frac{3}{5}\) + 4
= 4\(\frac{3}{5}\)

Write the correct mixed number in each box.
Question 12.

Answer:

Explanation:
1 + \(\frac{4}{7}\) = 1 \(\frac{4}{7}\)
2 + \(\frac{6}{7}\) = 2\(\frac{6}{7}\)

Write a mixed number for each item.
Question 13.
The pears have a weight of

Answer:

Explanation:
Weight of the pears = 1 + \(\frac{2}{4}\)
= 1 \(\frac{1}{2}\) pounds.

Question 14.
The worm started crawling from 0 centimeters.
It has crawled _____ centimeters.

Answer:
It has crawled 7\(\frac{7}{1}\) or 7.7 centimeters.

Explanation:
Number of centimeters the worm travelled = 0 + 7\(\frac{14}{2}\)
= 7\(\frac{7}{1}\) or 7.7 centimeters.

Write each mixed number in simplest form.
Example

Question 15.

Answer:

Explanation:
2\(\frac{4}{6}\) = 2\(\frac{2}{3}\)

Question 16.
3\(\frac{4}{8}\) = ____
Answer:
3\(\frac{4}{8}\) = 3\(\frac{1}{2}\)

Explanation:
3\(\frac{4}{8}\) = 3\(\frac{1}{2}\)

Question 17.
5\(\frac{6}{9}\) = ____
Answer:
5\(\frac{6}{9}\) = 5\(\frac{2}{3}\)

Explanation:
5\(\frac{6}{9}\) = 5\(\frac{2}{3}\)

Question 18.
6\(\frac{4}{12}\) = ____
Answer:
6\(\frac{4}{12}\) = 6\(\frac{1}{3}\)

Explanation:
6\(\frac{4}{12}\) = 6\(\frac{1}{3}\)

Question 19.
4\(\frac{3}{6}\) = ____
Answer:
4\(\frac{3}{6}\) = 4\(\frac{1}{2}\)

Explanation:
4\(\frac{3}{6}\) = 4\(\frac{1}{2}\)

Write each fraction and mixed number in a box to show its correct location on the number line.
Question 20.
1\(\frac{1}{2}\)
Answer:

Explanation:
1\(\frac{1}{2}\) = (2 + 1) ÷ 2
= 3 ÷ 2
= 1.5.

Question 21.
\(\frac{1}{2}\)
Answer:

Explanation:
\(\frac{1}{2}\) = 0.5.

Question 22.
1\(\frac{3}{4}\)
Answer:

Explanation:
1\(\frac{3}{4}\) = (4 + 3) ÷ 4
= 7 ÷ 4
= 1.75.

Fill in the boxes with fractions or mixed numbers. Express each answer in simplest form.
Example

Question 23.

Answer:

Explanation:
2 + \(\frac{1}{2}\)
= 2\(\frac{2}{4}\)
= 2\(\frac{1}{2}\)

3 + \(\frac{3}{4}\)
= 3 \(\frac{3}{4}\)

Question 24.

Answer:

Explanation:
3 +  \(\frac{1}{4}\)
= 3 \(\frac{1}{4}\)
4 + \(\frac{1}{2}\)
= 4 \(\frac{1}{2}\)

Question 25.

Answer:

Explanation:
5 +  \(\frac{1}{2}\)
= 5 \(\frac{1}{2}\)
5 + \(\frac{3}{5}\)
= 5 \(\frac{3}{5}\)

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