Holt CA Course 1
8-5 Disjoint Events
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Holt CA Course 1
8-5 Disjoint Events
Warm Up A bag contains 18 marbles; 4 are green, 8 are red, and 6 are yellow. Find the probability of each event when a marble is chosen at random.
1. yellow 2. red
3. green 4. blue
5. not red
13
0
49
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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
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
Vocabularydisjoint events
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
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
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) = 1
11P(A or E) = P(A) + P(E) Add the probabilities of
the individual events. = + =2
11111
311
311The probability of choosing an A or an E is .
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)
= + = =136
536
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
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
Seco
nd 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
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
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) = 3
10P(I or E) = P(I) + P(E) Add the probabilities of
the individual events. = + = =1
10310
410
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 eight equal sectors numbered 1-6
P(2) = 16 P(4) = 1
6P(2 or 4) = P(2) + P(4) Add the probabilities of
the individual events. = + = =1
616
26
13The probability of choosing a 2 or a 4 is .
13
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
Seco
nd 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) = 3
36P(sum of 3 or sum of 4) = P(sum of 3) + P(sum of 4)
= + = 236
336
536 5
36The 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?
Holt CA Course 1
8-5 Disjoint EventsLesson Quiz: Part I
1. Determine whether choosing a fiction book or a nonfiction book is a set of disjoint events.
Explain.
Find the probability of each set of disjoint events.2. Choosing an O or an E from the letters in outcome
3. Rolling an odd number or a 6 on a number cube
Disjoint; you cannot choose a book that is both fiction and nonfiction.
3 7
2 3
Holt CA Course 1
8-5 Disjoint Events
Lesson Quiz: Part II
4. David rolls two number cubes at the same time. If the product of the numbers rolled is 25 or 30, he will win the game. Use a grid to find the sample space. The find the probability that David will win the game.
1 12
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 1-8P(3) = 1
8 P(4) = 18
P(3 or 4) = P(3) + P(4) Add the probabilities of the individual events.
= + = =18
18
28
14The probability of choosing a 3 or a 4 is .
14
Additional Example 2: Finding the Probability of Disjoint Events