Download Holt CA Course 1

8-5 Disjoint Events
Preview
Warm Up
California Standards
Lesson Presentation
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 1
3
2. red 4
9
2
9
4. blue 0
3. green
5. not red 5
9
Holt CA Course 1
8-5 Disjoint Events
California
Standards
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
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
Probability of Two Disjoint Events
P(A or B) = P(A) + P(B)
Probability of
either event
Probability
Probability of
of one event other event
Reading Math
Disjoint events are sometimes called
mutually exclusive events.
Holt CA Course 1
8-5 Disjoint Events
Additional Example 1: Identifying Disjoint Events
Determine whether each set of events is
disjoint. Explain.
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
Additional Example 2: Finding the Probability of
Disjoint Events
Find the probability of each set of disjoint
events.
A. choosing an A or an E from the letters in
the word mathematics
P(E) = 1
P(A) = 2
11
11
P(A or E) = P(A) + P(E)
2
1
3
= 11 + 11= 11
Add the probabilities of
the individual events.
The probability of choosing an A or an E is 3 .
11
Holt CA Course 1
8-5 Disjoint Events
Additional Example 3: Recreation Application
Sharon rolls two number cubes. What is the
probability that the sum of the numbers shown
on the cubes is 2 or 8?
Step 1 Use a grid to find the sample space.
Second Roll
First Roll
1
2
3
4
5
6
1
2
3
4
5
6
7
Holt CA Course 1
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
The grid shows all
possible sums.
There are 36 equally likely
outcomes in the sample
space.
8-5 Disjoint Events
Additional Example 3 Continued
Sharon rolls two number cubes. What is the
probability that the sum of the numbers shown
on the cubes is 2 or 8?
Step 2 Find the probability of the set of disjoint
events.
1
P(sum of 2) = 36
5
36
P(sum of 2 or sum of 8) = P(sum of 2) + P(sum of 8)
P(sum of 8) =
5
1
1
6
= 36 + 36 = 36 = 6
1
The probability that the sum of the cubes is 2 or 8 is 6 .
Holt CA Course 1
8-5 Disjoint Events
Check It Out! Example 1
Determine whether each set of events is
disjoint. Explain.
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 soup
The 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
Check It Out! Example 2
Find the probability of each set of disjoint
events.
A. choosing an I or an E from the letters in the
word centimeter
P(I) =
1
10
P(E) = 3
10
P(I or E) = P(I) + P(E)
1
3
4
2
= 10 + 10= 10 = 5
Add the probabilities of
the individual events.
The probability of choosing an I or an E is 2 .
5
Holt CA Course 1
8-5 Disjoint Events
Check It Out! Example 2
Find the probability of each set of disjoint
events.
B. spinning a 2 or a 4 on a spinner with eight
equal sectors numbered 1-6
P(2) = 1
6
P(4) = 1
6
P(2 or 4) = P(2) + P(4)
1
1
2
1
= 6 + 6= 6= 3
Add the probabilities of
the individual events.
1
The probability of choosing a 2 or a 4 is 3 .
Holt CA Course 1
8-5 Disjoint Events
Check It Out! Example 3
Sun Li rolls two number cubes. What is the
probability that the sum of the numbers shown
on the cubes is 3 or 4?
Step 1 Use a grid to find the sample space.
Second Roll
First Roll
1
2
3
4
5
6
1
2
3
4
5
6
7
Holt CA Course 1
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
The grid shows all
possible sums.
There are 36 equally likely
outcomes in the sample
space.
8-5 Disjoint Events
Check It Out! Example 3 Continued
Sun Li rolls two number cubes. What is the
probability that the sum of the numbers shown
on the cubes is 3 or 4?
Step 2 Find the probability of the set of disjoint
events.
P(sum of 3) = 2
P(sum of 4) = 3
36
36
P(sum of 3 or sum of 4) = P(sum of 3) + P(sum of 4)
= 2 + 3 = 5
36 36 36
5
The probability that the sum of the cubes is 3 or 4 is 36.
Holt CA Course 1
8-5 Disjoint Events
Lesson Quiz: Part I
1. Determine whether choosing a fiction book or a
nonfiction book is a set of disjoint events.
Explain.
Disjoint; you cannot choose a book that is both
fiction and nonfiction.
Find the probability of each set of disjoint
events.
2. Choosing an O or an E from the letters in
outcome
3
7
3. Rolling an odd number or a 6 on a number 2
3
cube
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