Holt CA Course 1

8-5 Disjoint Events

SDAP3.4 Understand that the probability of either of two disjoint events occurring is the sum of the two individual probabilities and that the probability of one event following another, in independent trials, is the product of the two probabilities.Also covered: SDAP3.1

California Standards

Holt CA Course 1

8-5 Disjoint Events

Vocabulary

disjoint events

Holt CA Course 1

8-5 Disjoint Events

On a game show, the letters in the word Hollywood are printed on cards and shuffled. A contestant will win a trip to Hollywood if the first card she chooses is printed with an O or an L.

Choosing an O or an L on the first card is an example of a set of disjoint events. Disjoint events cannot occur in the same trial of an experiment.

Holt CA Course 1

8-5 Disjoint Events

Determine whether each set of events is disjoint. Explain.

Additional Example 1: Identifying Disjoint Events

A. choosing a dog or a poodle from the animals at an animal shelter

The event is not disjoint. A poodle is a type of dog, so it is possible to choose an animal that is both a dog and a poodle.

B. choosing a fish or a snake from the animals at a pet store

The event is disjoint. Fish and snakes are different types of animals, so you cannot choose an animal that is both a fish and a snake.

Holt CA Course 1

8-5 Disjoint Events

Determine whether each set of events is disjoint. Explain.

Check It Out! Example 1

A. choosing a bowl of soup or a bowl of chicken noodle soup from the cafeteria

The event is not disjoint. Chicken noodle is a type of soup, so it is possible to choose a bowl of chicken noodle soup and soup.

B. choosing a bowl of chicken noodle soup or broccoli cheese soupThe event is disjoint. Chicken noodle and broccoli cheese are different types of soups, so you cannot choose a soup that is both chicken noodle and broccoli cheese.

Holt CA Course 1

8-5 Disjoint Events

Probability of Two Disjoint EventsProbability of Two Disjoint Events

PP((AA or or BB) = ) = PP((AA) + ) + PP((BB))

Probability of either event

Probability of one event

Probability of other event

Disjoint events are sometimes called mutually exclusive events.

Reading Math

Holt CA Course 1

8-5 Disjoint Events

Find the probability of each set of disjoint events.

Additional Example 2: Finding the Probability of Disjoint Events

A. choosing an A or an E from the letters in the word mathematics

P(A) = 211

P(E) = 111

P(A or E) = P(A) + P(E)Add the probabilities of the individual events.

= + =211

111

311

311

The probability of choosing an A or an E is .

Holt CA Course 1

8-5 Disjoint Events

Find the probability of each set of disjoint events.

B. spinning a 3 or a 4 on a spinner with eight equal sectors numbered 1-8

P(3) = 18

P(4) = 18

P(3 or 4) = P(3) + P(4) Add the probabilities of the individual events.

= + = =18

18

28

14

The probability of choosing a 3 or a 4 is .

14

Additional Example 2: Finding the Probability of Disjoint Events

Holt CA Course 1

8-5 Disjoint Events

Find the probability of each set of disjoint events.

Check It Out! Example 2

A. choosing an I or an E from the letters in the word centimeter

P(I) = 110

P(E) = 310

P(I or E) = P(I) + P(E) Add the probabilities of the individual events.

= + = =1

103

10410

25The probability of choosing an I or an E is .

25

Holt CA Course 1

8-5 Disjoint Events

Find the probability of each set of disjoint events.

Check It Out! Example 2

B. spinning a 2 or a 4 on a spinner with six equal sectors numbered 1-6

P(2) = 16

P(4) = 16

P(2 or 4) = P(2) + P(4) Add the probabilities of the individual events.

= + = =16

16

26

13The probability of choosing a 2 or a 4 is .

13

Holt CA Course 1

8-5 Disjoint Events

Sharon rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 2 or 8?

Additional Example 3: Recreation Application

Step 1 Use a grid to find the sample space.

1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

First Roll

Secon

d R

oll

The grid shows all possible sums.

There are 36 equally likely outcomes in the sample space.

Holt CA Course 1

8-5 Disjoint Events

P(sum of 2) = 136 P(sum of 8) = 5

36P(sum of 2 or sum of 8) = P(sum of 2) + P(sum of 8)

= + = =1

36536

636

16The probability that the sum of the cubes is 2 or 8 is .

16

Sharon rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 2 or 8?

Additional Example 3 Continued

Step 2 Find the probability of the set of disjoint events.

Holt CA Course 1

8-5 Disjoint Events

Sun Li rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 3 or 4?

Check It Out! Example 3

Step 1 Use a grid to find the sample space.

1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

First Roll

Secon

d R

oll

The grid shows all possible sums.

There are 36 equally likely outcomes in the sample space.

Holt CA Course 1

8-5 Disjoint Events

P(sum of 3) = 236

P(sum of 4) = 336

P(sum of 3 or sum of 4) = P(sum of 3) + P(sum of 4)

= + = 236

336

536

536The probability that the sum of the cubes is 3 or 4 is .

Check It Out! Example 3 Continued

Step 2 Find the probability of the set of disjoint events.

Sun Li rolls two number cubes. What is the probability that the sum of the numbers shown on the cubes is 3 or 4?