﻿ Experimental and Theoretical Probability Worksheets | Problems & Solutions
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Experimental and Theoretical Probability Worksheets
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1.
Sheela has a collection of 89 triangles in her bag. She sorted the triangles into pattern of 6. The leftover triangles were put back into the bag. How many patterns were made? How many triangles were put back in the bag? a. 14 patterns; 5 triangles put back b. 14 patterns; 2 triangles put back c. 12 patterns; 2 triangles put back d. 15 patterns; 1 triangles put back

#### Answer: (a)

Correct answer : (1)
2.
Frank used 5 squares to make each building. How many buildings did he make with 57 squares? How many squares were not used in the building? a. 12 buildings; 2 squares not used b. 10 buildings; 7 squares not used c. 11 buildings; 2 squares not used d. 12 buildings; 3 squares not used

#### Answer: (c)

Correct answer : (3)
3.
Frank used 5 squares to make each building. How many buildings did he make with 57 squares? How many squares were not used in the building? a. 12 buildings; 2 squares not used b. 10 buildings; 7 squares not used c. 12 buildings; 3 squares not used d. 11 buildings; 2 squares not used

#### Answer: (d)

Correct answer : (4)
4.
Sheela had a collection of 89 triangles in her bag. She sorted the triangles into pattern of 6. The leftover triangles were put back into the bag. How many patterns were made? How many triangles were put back in the bag? a. 14 patterns; 5 triangles put back b. 14 patterns; 2 triangles put back c. 12 patterns; 2 triangles put back d. 15 patterns; 1 triangles put back

#### Answer: (a)

Correct answer : (1)
5.
Frank used 5 squares to make each building. How many buildings did he make with 57 squares? How many squares were not used in the building? a. 12 buildings; 3 squares not used b. 11 buildings; 2 squares not used c. 12 buildings; 2 squares not used d. 10 buildings; 7 squares not used

#### Answer: (b)

Correct answer : (2)
6.
Jack collected 2 blue, 1 red, and 2 green marbles. Which list shows the possible combination of 4 marbles picked at the same time? a. 2 blue, 2 green, and 1 red b. 1 blue, 2 green, and 1 red c. 1 blue, 1 green, and 2 red d. 0 blue, 1 red, and 3 green

#### Solution:

In all, Jack collected 5 marbles.

Out of those 5, 4 marbles are to be picked.

1 blue, 1 green, and 2 red - this combination is not possible as there is only one red marble collected.

0 blue, 1 red, and 3 green - this combination is not possible as there are only two green marbles collected.

2 blue, 2 green, and 1 red - this combination adds up to 5 marbles, but we need to pick just 4 marbles.

1 blue, 2 green, and 1 red is the possible combination of 4 marbles picked at the same time.
[1 + 2 + 1 = 4.]

Correct answer : (2)
7.
Jake has a set of plane shapes as shown in the figure. Which of the following is the possible outcome if Jake chooses 6 shapes?  a. 4 rectangles, 1 square, and 1 circle b. 2 squares, 2 rectangles, and 2 circles c. 1 rectangle, 2 squares, 2 triangles, and 1 circle d. 2 squares, 3 rectangles, 1 triangle, and 1 circle

#### Solution:

The given figure has 2 triangles, 3 rectangles, 2 squares, and 1 circle.

Jake has to choose any of those 6 shapes.

4 rectangles, 1 square, and 1 circle is not a possible outcome.
[Since Jake has only 3 rectangles.]

1 rectangle, 2 squares, 2 triangles, and 1 circle is a possible outcome.
[Since all these 6 shapes are available with Jack.]

2 squares, 3 rectangles, 1 triangle, and 1 circle is not a possible outcome.
[Since 6 shapes are required but 3 + 1 + 1 = 5.]

2 squares, 2 rectangles, and 2 circles is not a possible outcome.
[Since Jake has only 1 circle.]

Among the choices "1 rectangle, 2 squares, 2 triangles, and 1 circle" is the possible outcome.

Correct answer : (3)
8.
Tim tossed a coin 28 times and recorded the results. He tossed heads 12 times. For what fraction of the tosses did the coin land heads up? a. $\frac{1}{3}$ b. $\frac{1}{7}$ c. $\frac{3}{7}$ d. $\frac{4}{7}$

#### Solution:

Experimental Probability(head) = Number of headsTotal number of tosses

P(heads) = 12 / 28
[Substitute the values.]

= 3 / 7
[Simplify.]

The experimental probability that the result is heads is 3 / 7.

Correct answer : (3)
9.
An employee of the ABC Toy Manufacturing Company checked 336 toys and found 14 of them defective. What is the experimental probability of choosing a defective toy? a. $\frac{1}{8}$ b. $\frac{5}{24}$ c. $\frac{1}{12}$ d. $\frac{1}{24}$

#### Solution:

Experimental probability = Number of defective toysTotal number of toys checked

P (defective) = 14 / 336
[Substitute the values.]

= 1 / 24
[Simplify.]

The probability of choosing a defective toy is 1 / 24.

Correct answer : (4)
10.
Madison High School ordered 47 new soccer balls. The students found 7 of the balls defective. What is the experimental probability of finding a defective ball? a. $\frac{7}{47}$ b. $\frac{1}{7}$ c. $\frac{2}{47}$ d. $\frac{1}{47}$

#### Solution:

Total number of balls = 47

Number of defective balls = 7

P (defective ball) = 7 / 47
[Number of defective ballsTotal number of balls.]

The probability of finding a defective ball is 7 / 47.

Correct answer : (1)

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