﻿ Independent and Dependent Events Worksheet | Problems & Solutions

# Independent and Dependent Events Worksheet

Independent and Dependent Events Worksheet
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1.
What is the probability of getting a composite number on the first roll of a die and getting a prime number on the second roll?
 a. $\frac{1}{2}$ b. $\frac{1}{6}$ c. $\frac{1}{4}$ d. $\frac{1}{3}$

#### Solution:

The two events of getting an composite number on first roll and getting a prime number on the second roll are independent events.

P(getting a composite number) = Number of composite numbers in a dieTotal number of outcomes = 3 / 6 = 1 / 2
[Composite numbers in a die are 1, 4 and 6.]

P(getting a prime number) = Number of prime numbers in a dieTotal number of outcomes = 3 / 6 = 1 / 2
[Primes in a die are 2, 3 and 5.]

P (getting a composite number on the first roll and a prime number on the second roll) = P(getting a composite number) × P(getting a prime number)

= 1 / 2 × 1 / 2 = 1 / 4
[Substitute and simplify.]

So, the probability of getting a composite number on the first roll and a prime number on the second roll is 1 / 4.

2.
There are 6 red marbles, 5 green marbles, and 4 yellow marbles in a bag. If Joe picks 2 marbles one after the other without replacement, then what is the probability that both are red in color?
 a. $\frac{2}{5}$ b. $\frac{1}{21}$ c. $\frac{4}{25}$ d. $\frac{1}{7}$

#### Solution:

P(picking red colored marble first) = Number of red colored marblesTotal number of marbles = 6 / 15

Since the first marble selected is not replaced back, the total number of marbles is reduced by one.

P(picking red colored marble next) = Number of red colored marblesTotal number of marbles = 5 / 14

P(picking red colored marble first, picking red colored marble next) = P(picking red colored marble first) × P(picking red colored marble next)

= 6 / 15 × 5 / 14 = 1 / 7
[Substitute and simplify the product.]

The probability that both are red in color is 1 / 7.

3.
There are 5 red roses, 3 yellow roses, and 8 white roses in a tray. If Stephanie picked 2 roses one after the other without replacing, then what is the probability of picking a white rose first and a red rose next?
 a. $\frac{1}{6}$ b. $\frac{5}{6}$ c. $\frac{1}{3}$ d. $\frac{2}{3}$

#### Solution:

P(picking a white rose first) = Number of white rosesTotal number of roses = 8 / 16

P(picking a red rose next) = Number of red rosesTotal number of roses = 5 / 15
[One white rose is already picked, so total number of roses is decreased by 1.]

P(picking a white rose first and a red rose next) = P(picking a white rose first) × P(picking a red rose next)

= 8 / 16 × 5 / 15
[Substitute the values.]

= 1 / 6
[Simplify.]

So, the probability of picking a white rose first and a red rose next is 1 / 6.

4.
There are 6 pink and 8 white balls in a bag. If two balls are drawn one after the other, then what is the probability of getting a pink ball first and white ball next, if the first ball drawn is replaced?
 a. $\frac{1}{7}$ b. $\frac{4}{7}$ c. $\frac{12}{49}$ d. $\frac{3}{7}$

#### Solution:

The two events, getting a pink ball first and a white ball next are independent events because the first ball drawn is replaced.

P (getting a pink ball first) = Number of pink ballsTotal number of balls = 6 / 14

P (getting a white ball next) = Number of white ballsTotal number of balls = 8 / 14

P (getting pink ball first and white next) = P (getting a pink ball first) × P (getting a white ball next) = 6 / 14 × 8 / 14 = 12 / 49
[Substitute and simplify the fraction.]

So, the probability of getting pink ball first and white ball next is 12 / 49.

5.
In a bag containing 15 billiard balls of three colors - red, yellow and white, Frank picks a ball at random, notes down the color and replaces it. He repeats this experiment 25 times and the results are shown in the table.
 Color Frequency Yellow 8 Red 13 White 4

Which is the best prediction of the number of balls of each color in the bag?
 a. 6 yellow, 9 red, 1 white b. 2 yellow, 5 red, 8 white c. 5 yellow, 8 red, 2 white d. 5 yellow, 10 red, 5 white

#### Solution:

From the table, the most likely outcome is red and the least likely outcome is white.

The total number of billiard balls is 15.

So, the total number of predictions should be 15 in such a way that the most likely outcomes are red balls and the least likely outcomes are white balls.

Among the choices, 5 yellow, 8 red, 2 white is the correct choice.

6.
The spinner shown was spun 30 times and the outcomes were noted down. Which of the following could be the most possible outcomes?

 a. A - 4, B - 15, C - 8, D - 3 b. A - 10, B - 5, C - 8, D - 7 c. A - 9, B - 7, C - 8, D - 6 d. A - 5, B - 16, C - 9, D - 2

#### Solution:

The spinner contains 8 equal sectors of which four sectors contain letter B, two sectors contain letter C, and the letters D and A have one sector each.

The most likely outcome is letter B.

The next most likely outcome is letter C.

The least likely outcomes are letter D and letter A.

As the spinner is spun 30 times, so the total number of outcomes should be 30 in such a way that the most likely outcome is letter B and the least likely outcomes are letter D and letter A.

Among the choices, A - 4, B - 15, C - 8, D - 3 is the correct choice.

7.
Which of the following events are not mutually exclusive?
 a. drawing a 3 and a diamond from a standard deck of cards b. rolling a 3 or 5 on a single roll of a number cube c. rolling a number greater than 3 or a multiple of 3 when a pair of dice is rolled d. tossing a coin and rolling a number cube

8.
Which of the following events are not mutually exclusive?
 a. drawing a 3 and a diamond from a standard deck of cards b. rolling a 3 or 5 on a single roll of a number cube c. rolling a number greater than 3 or a multiple of 3 when a pair of dice is rolled d. tossing a coin and rolling a number cube

9.
4 red cubes and 4 white cubes are there in a basket. If two cubes are drawn at random, then what is the probability that the first one is white and the second one is red?
 a. $\frac{1}{2}$ b. $\frac{5}{7}$ c. $\frac{2}{7}$ d. $\frac{1}{7}$

#### Solution:

The two events are dependent events as the second cube is drawn without replacing the first cube.

The two events are dependent events as the second cube is drawn without replacing the first cube.

P(drawing a white cube first) = 4 / 8

P(drawing a red cube first) = 10 / 20

P(drawing a red cube next) = 4 / 7

P(drawing a black cube next) = 10 / 19

P(drawing a white cube first and red cube next) = P(drawing a white cube first) × P(drawing a red cube next)

P(drawing a red cube first and black cube next) = P(drawing a red cube first) × P(drawing a black cube next)

= 48 × 47 = 27
[Substitute and simplify.]

= 1020 × 1019 = 519
[Substitute and simplify.]

Probability that the first one is white and the second one is red is 2 / 7.

Probability that the first one is red and the second one is black is 5 / 19.

10.
Arthur has 4 brown, 5 violet, and 3 pink sweaters. If he selects 2 sweaters one after the other without replacement, then what is the probability that both are violet in color?
 a. $\frac{20}{133}$ b. $\frac{7}{44}$ c. $\frac{5}{33}$ d. None of the above

#### Solution:

P(choosing a violet sweater first) = Number of violet sweatersTotal number of sweaters = 5 / 12

Since the first sweater selected is not replaced back, the total number of sweaters is reduced by one.

The two events are dependent events.

P(choosing a violet sweater next) = Number of violet sweatersTotal number of sweaters = 4 / 11

P(violet colored sweater first, violet colored sweater next) = P(violet colored sweater first) × P(violet colored sweater next)

= 5 / 12 × 4 / 11 = 5 / 33
[Substitute and simplify the product.]

The probability that both are violet in color is 5 / 33.