Download Dependent Events

LESSON
20.3
Name
Dependent Events
Class
Date
20.3 Dependent Events
Essential Question: How do you find the probability of dependent events?
Common Core Math Standards
The student is expected to:
Resource
Locker
S-CP.8
Explore
Apply the general Multiplication Rule in a uniform probability model,
P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms
of the model. Also S-CP.3, S-CP.4, S-CP.5
You know two tests for the independence of events A and B:
Mathematical Practices
1.
2.
MP.6 Precision
If P(A∣B) = P(A), then A and B are independent.
If P(A ∩ B) = P(A) ∙ P(B), then A and B are independent.
Two events that fail either of these tests are dependent events because the occurrence of one
event affects the occurrence of the other event.
Language Objective
Work with a partner to brainstorm examples of dependent events.
A
You can use the Multiplication Rule to find the
probability of dependent events.
PREVIEW: LESSON
PERFORMANCE TASK
View the Engage section online. Discuss the
photograph. Ask students to describe the sport that is
pictured. Then preview the Lesson Performance Task.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Senior
Style/Getty Images
ENGAGE
Essential Question: How do you find
the probability of dependent events?
Finding a Way to Calculate the Probability of
Dependent Events
The two-way frequency table shows the results of a survey
of 100 people who regularly walk for exercise. Let O be the
event that a person prefers walking outdoors. Let M be the
event that a person is male. Find P(O), P(M), and P(O ∩ M)
as fractions. Then determine whether events O and M are
independent or dependent.
Prefers walking
outdoors
Prefers walking on
a treadmill
Total
Male
40
10
50
Female
20
30
50
Total
60
40
100
60
3 ( ) ___
50
40
1
2
P(O) = ___
=_
, P M = 100
=_
. and P(O ∩ M) = ___
=_
.
5
5
100
2
100
3 _
3
Since P(O) · P(M) = _
∙ 1 = __
≠ P(O ∩ M), events O and M are dependent.
5 2
10
B
Calculate the conditional probabilities P(O∣M) and P(M∣O).
40
4
n(O ∩ M)
P(O∣M) = _ = _ = _
5
n(M)
50
40
2
n(O ∩ M)
P(M∣O) = _ = _ = _
3
(
)
nO
60
Module 20
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Lesson 3
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A2_MNLESE385900_U8M20L3 1031
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Lesson 20.3
30
= 5.
t.
_1 . and P(O ∩ M) = 100
50
dependen
___
and M are
3 , P(M) = 100 = 2
60 = _
___
events O
3 ≠ P(O ∩ M),
P(O) = 100 5
_3 ∙ _1 = __
10
) · P(M) = 5 2
Since P(O
P(M∣O).
O∣M) and
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ional proba
condit
the
4
Calculate
40
_
)
n(O ∩ M = _ = 5
_
50
P(O∣M) = n(M)
Module 20
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10
20
Total
A2_MNLE
ng on
Prefers walki
a treadmill
40
2
40
_
)
n(O ∩ M = _ = 3
_
60
n(O)
Lesson 3
1031
8/26/14
12:21 PM
8/26/14 12:19 PM

Complete the multiplication table using the fractions for P(O) and P(M) from Step A and
the fractions for P(O∣M) and P(M∣O) from Step B.
x
( )
P(M∣O)
P O∣M

P(O)
12
_3 ∙ _4 = __
5 5
25
_3 ∙ _2 = _2
5
3
5
EXPLORE
P(M)
_1 ∙ _4 = _2
5
2 5
_1 ∙ _2 = _1
2
5
Finding a Way to Calculate the
Probability of Dependent Events
5
Do any of the four products in Step C equal P(O ∩ M), calculated in Step A? If so, which of
the four products?
INTEGRATE TECHNOLOGY
2
Yes, the products P(O) ∙ P(M∣O) and P(M) ∙ P(O|M) both equal _
, the value of P(O ∩ M).
5
Students have the option of doing the Explore activity
either in the book or online.
Reflect
1.
P(A ∩ B)
In a previous lesson you learned the conditional probability formula P(B∣A) = ______. How does this
P(A)
formula explain the results you obtained in Step D?
(A ⋂ B)
P
______
by P(A) gives P(A) ∙ P(B|A) = P(A ⋂ B).
Multiplying both sides of P(B|A) =
P(A)
Letting A = O and B = M gives P(O) ∙ P(M|O) = P(O ⋂ M), while letting A = M and B = O
QUESTIONING STRATEGIES
Suppose two events A and B are dependent.
How are their probabilities related? If events
A and B are dependent events, P(A and B) =
P(A|B) ⋅ P(B) or P(A and B) = P(B|A) ⋅ P(A).
gives P(M) ∙ P(O|M) = P(M ⋂ O) = P(O ⋂ M),
2.
Let F be the event that a person is female. Let T be the event that a person prefers walking on a treadmill.
Write two formulas you can use to calculate P(F ⋂ T). Use either one to find the value of P(F ⋂ T), and
then confirm the result by finding P(F ⋂ T) directly from the two-way frequency table.
P(F ⋂ T) = P(F) ∙ P(T|F); P(F ⋂ T) = P(T ⋂ F) = P(T) ∙ P(F|T). Using the first of these
50 __
3
1 _
formulas gives P(F ⋂ T) = ___
∙ 30 = _
∙ 3 = __
. Calculating P(T ⋂ F) directly from the
100 50
2 5
10
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Point out that if two events occur with
30
3
table gives ___
= __
.
100
10
Explain 1
Finding the Probability of Two Dependent Events
Multiplication Rule
P(A ⋂ B) = P(A) ∙ P(B|A) where P(B|A) is the conditional probability of event B,
given that event A has occurred.
There are 5 tiles with the letters A, B, C, D, and E in a bag. You
choose a tile without looking, put it aside, and then choose
another tile without looking. Use the Multiplication Rule to find
the specified probability, writing it as a fraction.
A C E
D
B
Example 1
replacement, then the events may not be dependent.
For example, if two marbles are drawn from a bag,
the events of drawing a marble are dependent. But, if
the first marble is replaced before the second one is
drawn, then the events become independent, since
the first event does not affect the second.
© Houghton Mifflin Harcourt Publishing Company
You can use the Multiplication Rule to find the probability of dependent events.
EXPLAIN 1
Finding the Probability of Two
Dependent Events
Module 20
1032
Lesson 3
PROFESSIONAL DEVELOPMENT
A2_MNLESE385900_U8M20L3 1032
Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.6,
which calls for students to “communicate with precision.” Students are already
familiar with some aspects of probability but, in this lesson, they must analyze
whether events are dependent or independent before they find the probabilities of
the events. They use multiple representations, including formulas, two-way tables,
and tree diagrams, to determine a probability. Students may also make an
organized list of all outcomes.
QUESTIONING STRATEGIES
8/26/14 12:25 PM
Why is probability of dependent events
different from the probability of independent
events? The occurrence of one event affects the
probability of the second event.
Dependent Events 1032
A
AVOID COMMON ERRORS
Find the probability that you choose a vowel followed by a consonant.
Let V be the event that the first tile is a vowel. Let C be the event that the second
2.
tile is a consonant. Of the 5 tiles, there are 2 vowels, so P(V) = _
5
3.
Of the 4 remaining tiles, there are 3 consonants, so P(C|V) = _
4
3.
3 =_
6 =_
2
By the Multiplication Rule, P(V ⋂ C) = P(V) ∙ P(V|C) = _ ∙ _
5 4
20
10
Help students who are confused about identifying
dependent events by having them communicate their
results from this lesson orally and in writing. Ask
them to check that they are using terms correctly. In
particular, be sure students understand that the word
dependent has a specific meaning in probability
theory that may be different from its meaning in
everyday conversation.
B
Find the probability that you choose a vowel followed by another vowel.
Let V1 be the event that the first tile is a vowel. Let V2 be the event that the second
2
tile is also a vowel. Of the 5 tiles, there are 2 vowels, so P(V1) =____
.
5
1
Of the 4 remaining tiles, there is 1 vowel, so P(V2|V1) =____
.
4
1
2
2
1
∙ ____
=____
=____
.
By the Multiplication Rule, P(V1 ⋂ V2) = P(V1) ∙ P(V2|V1) =____
5
4
20
10
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Point out the two different ways to determine
Your Turn
A bag holds 4 white marbles and 2 blue marbles. You choose a marble without
looking, put it aside, and choose another marble without looking. Use the
Multiplication Rule to find the specified probability, writing it as a fraction.
the probability of dependent events: (1) by using the
formula for the multiplication rule, and (2) by
making a tree diagram. You may also wish to tell
students that they can determine the probability by
thinking about permutations to find the number of
outcomes in the sample space and the number of
favorable outcomes.
3.
Find the probability that you choose a white marble followed by a blue marble.
Let W be the event that the first marble is white. Let B be the event that the second
marble is blue.
4
2
P(W) = _
=_
6
3
2
P(B|W) = _
5
© Houghton Mifflin Harcourt Publishing Company
4
2 _
P(W ⋂ B) = P(W) ∙ P(B|W) = __
∙ 2 = __
15
3 5
4.
Find the probability that you choose a white marble followed by another white marble.
Let W1 be the event that the first marble is a white marble. Let W2 be the event that the
second marble is also a white marble.
4
2
=_
P(W1) = _
6
3
3
P(W2|W1) = _
5
2
2 _
P(W1 ⋂ W2) = P(W1) ∙ P(W2|W1) = __
∙3=_
5
3 5
Module 20
1033
Lesson 3
COLLABORATIVE LEARNING
A2_MNLESE385900_U8M20L3 1033
Peer-to-Peer Activity
Have students work in pairs. Give each pair a set of 5 to 10 playing cards or
numbered index cards. Have students lay out the cards face up, write a list of
possible independent and dependent events using the cards, and find the
probabilities. Have one student identify the favorable outcomes and the other
student identify the sample space for each event.
1033
Lesson 20.3
8/27/14 5:12 PM
Explain 2
Finding the Probability of Three or More
Dependent Events
EXPLAIN 2
You can extend the Multiplication Rule to three or more events. For instance, for three events A, B, and C, the rule
becomes P(A ⋂ B ⋂ C) = P(A) ∙ P(B|A) ∙ P(C|A ⋂ B).
Example 2

Finding the Probability of Three or
More Dependent Events
You have a key ring with 7 different keys. You’re attempting to unlock a door in
the dark, so you try keys one at a time and keep track of which ones you try.
Find the probability that the third key you try is the right one.
QUESTIONING STRATEGIES
Let W1 be the event that the first key you try is wrong. Let W2 be the event that the second key you try is
also wrong. Let R be the event that the third key you try is right.
When does replacement affect
independence? When there is replacement,
the events are independent because the outcome of
the first event does not affect the probability of the
second event.
6.
On the first try, there are 6 wrong keys among the 7 keys, so P(W1) = _
7
5.
On the second try, there are 5 wrong keys among the 6 remaining keys, so P(W2|W1) = _
6
1.
On the third try, there is 1 right key among the 5 remaining keys, so P(R|W2 ⋂ W1) = _
5
6 ∙_
5 ∙_
1 =_
1.
By the Multiplication Rule, P(W1 ⋂ W2 ⋂ R) = P(W1) ∙ P(W2|W1) ∙ P(R|W1 ⋂ W2) = _
7 6 5
7

Why can you use the multiplication rule
P(A and B) = P(A) ⋅ P (B|A) for both
dependent and independent events? If the events are
independent, P (B|A) is equal to P(B).
Find the probability that one of the first three keys you try is right.
There are two ways to approach this problem:
1. You can break the problem into three cases: (1) the first key you try is right; (2) the
first key is wrong, but the second key is right; and (3) the first two keys are wrong, but
the third key is right.
2. You can use the complement: The complement of the event that one of the first three
keys is right is the event that none of the first three keys is right.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Ask students if they think that the
Use the second approach.
Let W1, W2, and W3 be the events that the first, second, and third keys, respectively, are wrong.
4
On the third try, there are 4 wrong keys among the 5 remaining keys, so P(W3|W2 ⋂ W1) = _.
5
By the Multiplication Rule,
6
5
4
4
P(W1 ⋂ W2 ⋂ W3) = P(W1) ∙ P(W2|W1) ∙ P(W3|W1 ⋂ W2) = _ ∙ _ ∙ _ = _.
7
5
6
7
The event W1 ⋂ W2 ⋂ W3 is the complement of the one you want. So, the probability that one of
4
3
the first three keys you try is right is 1 - P(W1 ⋂ W2 ⋂ W3) = 1 − _ = _.
7
7
Module 20
1034
© Houghton Mifflin Harcourt Publishing Company
6
5
From Part A, you already know that P(W1) = _ and P(W2|W1) = _.
7
6
multiplication rule for dependent events can be
extended, as it was for independent events, and
why. Yes; for example, a formula for three events is
P(A and B and C) = P(A) ⋅ P(B | A) ⋅ P(C | A and B).
Since each part of this formula can be found for
dependent events, you can extend the rule.
Lesson 3
DIFFERENTIATE INSTRUCTION
A2_MNLESE385900_U8M20L3 1034
Modeling
04/09/14 7:26 PM
Help students understand the difference between independent events and
dependent events by discussing choosing two colored marbles from a bag: (1) with
replacement and (2) without replacement. When you select a marble with
replacement, you select the first marble, note its color, put the marble back in the
bag, and then choose the second marble. In this case, the two selections are
independent events. Without replacement, you select the first marble, put it aside,
and then choose the second marble. In this case, the two selections are dependent
events because the marble you chose first changed the sample space for your
second selection.
Dependent Events 1034
Reflect
5.
In Part B, show that the first approach to solving the problem gives the same result.
Let W1 and W2 be the events that the first and second keys, respectively,
are wrong. Let R be the event that a key is right. Calculate three
probabilities: P(R), P(W1 ⋂ R), and P(W1 ⋂ W2 ⋂ R).
1
There is 1 right key among the 7 keys on the first try, so P(R) = _
.
7
There are 6 wrong keys among the 7 keys on the first try and then 1 right key among the
6 _
1
6 remaining keys on the second try, so P(W1 ⋂ R) = P(W1)∙P(R|W1) = _
∙1=_
.
7 6
7
1
From Part A, you know that P(W1 ⋂ W2 ⋂ R) = _
.
7
Because the three events R, W1 ⋂ R, and W2 ⋂ W1 ⋂ R are mutually exclusive, the
probability that one of the first three keys is right is the sum of the probabilities of those
3
1
1
1
events: P(R) + P(W1 ⋂ R) + P(W2 ⋂ W1 ⋂ R) = _
+_
+_
=_
.
7
7
7
7
6.
In Part A, suppose you don’t keep track of the keys as you try them. How does the probability change?
Explain.
6
If you don’t keep track of the keys, the probability of trying a wrong key is always _
, and
7
1
the probability of choosing the right key is always _
. The events W1, W2, and R are now
7
independent, so you can multiply their probabilities without considering conditional
6 _
36
1
probabilities: P(W1 ⋂ W2 ⋂ R) = P(W1)∙P(W2)∙P(R) = _
∙6∙_
= ___
7 7 7
343
Your Turn
© Houghton Mifflin Harcourt Publishing Company
Three people are standing in line at a car rental agency at an airport. Each person is
willing to take whatever rental car is offered. The agency has 4 white cars and 2 silver
ones available and offers them to customers on a random basis.
7.
Find the probability that all three customers get white cars.
Let W1, W2, and W3 be the events that the first, second, and third customers, respectively,
get a white car.
P(W1 ⋂ W2 ⋂ W3) = P(W1) ∙ P(W2|W1) ∙ P(W3|W1 ⋂ W2)
4 _
2
=_
∙3∙_
6 5 4
2 _
1
=_
∙3∙_
3 5 2
1
=_
5
1.
So, the probability that all three customers get a white car is _
5
Module 20
1035
Lesson 3
LANGUAGE SUPPORT
A2_MNLESE385900_U8M20L3 1035
Connect Vocabulary
Help students understand dependent events by having them create a poster with
examples of dependent events that may be familiar, and that show simple
calculations for the probability of the dependent events. You may want to prompt
students to highlight the words that correspond to the event that is assumed to
have already occurred and have them focus on the fact that this event affects the
outcome of the other event.
1035
Lesson 20.3
8/27/14 5:12 PM
8.
Find the probability that two of the customers get the silver cars and one gets a white car.
ELABORATE
Let W1 and S1 represent the events that the first customer gets a white or silver car,
respectively. Let W2 and S2 represent the events that the second customer gets a white
or silver car, respectively. Let W3 and S3 represent the events that the third customer gets
QUESTIONING STRATEGIES
a white or silver car, respectively. Calculate three probabilities: P(W1 ⋂ S2 ⋂ S3),
How is the multiplication rule used to find the
probability of dependent or independent
events A and B? Sample answer: If A and B are
dependent, then the rule is
P(A and B) = P(A) ⋅ P(B | A). If A and B are
independent, then P(B | A) can be replaced by P(B)
and the multiplication rule becomes
P(A and B) = P(A) ⋅ P(B).
P(S1 ⋂ W2 ⋂ S3), and P(S1 ⋂ S2 ⋂ W3).
1
_4 ∙ _2 ∙ _1 = _
15
6 5 4
1
2 4 1
P(S1 ⋂ W2 ⋂ S3) = P(S1) ∙ P(W2|S1) ∙ P(S3|S1 ⋂ W2) = _ ∙ _ ∙ _ = _
15
6 5 4
1
2 1 4
P(S1 ⋂ S2 ⋂ W3) = P(S1) ∙ P(S2|S1) ∙ P(W3|S1 ⋂ S2) = _ ∙ _ ∙ _ = _
P(W1 ⋂ S2 ⋂ S3) = P(W1) ∙ P(S2|W1) ∙ P(S3|W1 ⋂ S2) =
Because the events W1 ⋂ S2 ⋂ S3, S1 ⋂ W2 ⋂ S3, and
6
5
4
15
S1 ⋂ S2 ⋂ W3 are mutually exclusive, you can add their probabilities:
3
1
1
1
1
P(W1 ⋂ S2 ⋂ S3) + P(S1 ⋂ W2 ⋂ S3) + P(S1 ⋂ S2 ⋂ W3) =
+
+
=
=
15
15
15
15
5
1
So, the probability that two customers get the silver cars and one gets a white car is .
5
_ _ _ _ _
_
SUMMARIZE THE LESSON
Elaborate
9.
State the multiplication rule,
P(A and B) = P(A) ⋅ P(B | A), in your own
words. The probability that dependent events
A and B take place is equal to the product of the
probability that A takes place and the probability
that B takes place given that A has already occurred.
When are two events dependent?
Two events are dependent when the occurrence of one event affects the
occurrence of the other.
10. Suppose you are given a bag with 3 blue marbles and 2 red marbles, and you are asked to find the
probability of drawing 2 blue marbles by drawing one marble at a time and not replacing the first marble
drawn. Why does not replacing the first marble make these events dependent? What would make these
events independent? Explain.
The first marble you draw changes the sample space for your second draw. That is, the
of the event that you get a blue marble on the second draw. Replacing the first marble
drawn returns the sample space to its original state so that the probability of getting a
blue marble doesn’t change from one draw to the next, which means that the events are
independent.
11. Essential Question Check-In According to the Multiplication Rule, when finding P(A⋂B) for
dependent events A and B, you multiply P(A) by what?
You multiply P(A) by P(B| A).
Module 20
A2_MNLESE385900_U8M20L3 1036
1036
© Houghton Mifflin Harcourt Publishing Company
occurrence of the event that you get a blue marble on the first draw affects the occurrence
Lesson 3
8/26/14 12:43 PM
Dependent Events 1036
Evaluate: Homework and Practice
EVALUATE
1.
Town officials are considering a property tax increase to finance the building of a
new school. The two-way frequency table shows the results of a survey of 110 town
residents.
ASSIGNMENT GUIDE
Supports a
property tax
increase
Does not support
a property tax
increase
Total
Lives in a household
with children
50
20
70
Concepts and Skills
Practice
Lives in a household
without children
10
30
40
Explore
Finding a Way to Calculate the
Probability of Dependent Events
Exercise 1
Total
60
50
110
Example 1
Determining the Probability of Two
Dependent Events
Exercises 2–3
Example 2
Finding the Probability of Three or
More Dependent Events
Exercises 4–14
the difference between dependent and independent
events. Then have them explain to each other how to
find the probabilities associated with dependent and
independent events. Ask them to use examples in
their explanations.
a. Let C be the event that a person lives in a household with children. Let S be the
event that a person supports a property tax increase. Are the events C and S
independent or dependent? Explain.
6
5
70
60
50
7
_
= _, P (S) = _ = _, and P(C ∩ S) = _ = _.
11
11
11
110
110
110
55
5
42
7
6
Since P(S) ∙ P(C) = _ ∙ _ = _ and P (C ∩ S) = _ = _,
P(C) =
11 11
11
121
121
P (C) ∙ P(S) ≠ P(C ∩ S), so the events C and S are dependent.
b. Find P(C|S) and P(S|C). Which of these two conditional probabilities can you
multiply with P(C) to get P(C ∩ S)? Which of the two can you multiply with P(S)
to get P(C ∩ S)?
P(C|S) =
5
5
50 _
50
_
= and P(S|C) = _ = _.
60
6
Exercise
A2_MNLESE385900_U8M20L3 1037
Lesson 20.3
11
7
11
11
6
11
5
6 5
Multiplying P(S) and P(C|S) gives _ ∙ _ = _ = P(C ∩ S).
Module 20
1037
7
70
5
7 5
Multiplying P(C) and P(S|C) gives _ ∙ _ = _ = P(C ∩ S).
© Houghton Mifflin Harcourt Publishing Company
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Have students work in small groups to explain
• Online Homework
• Hints and Help
• Extra Practice
Lesson 3
1037
Depth of Knowledge (D.O.K.)
Mathematical Practices
1
2 Skills/Concepts
MP.4 Modeling
2–11
2 Skills/Concepts
MP.1 Problem Solving
12
2 Skills/Concepts
MP.4 Modeling
13
3 Strategic Thinking
MP.3 Logic
14
3 Strategic Thinking
MP.2 Reasoning
8/28/14 8:58 PM
2.
A mall surveyed 120 shoppers to find out whether they typically wait for a sale to get a
better price or make purchases on the spur of the moment regardless of price. The
two-way frequency table shows the results of the survey.
Waits for
a Sale
Buys on
Impulse
Total
Woman
40
10
50
Man
50
20
70
Total
90
30
120
AVOID COMMON ERRORS
When students start finding probabilities for events A
and B, it is easy to confuse the context. Have students
read each exercise and highlight the key words and
phrases. Suggest that they first decide whether the
given events are dependent or independent. Then
have them compute the probability and use it as a
guide to determine whether the answer is reasonable.
a. Let W be the event that a shopper is a woman. Let
S be the event that a shopper typically waits for a sale. Are the events W and S
independent or dependent? Explain.
5
3
50
90
40
1.
_
= _, P(S) = _ = _, and P(W ∩ S) = _ = _
4
120
12
120
120
3
5
3
5
1
_
_
_
_
≠ = P(W ∩ S), the events W and S are dependent.
· =
Since P(W) ∙ P(S) =
P(W) =
12
4
16
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Tree diagrams are useful for understanding
3
b. Find P(W|S) and P(S|W) . Which of these two conditional probabilities can you
multiply with P(W) to get P(W ∩ S) ? Which of the two can you multiply with P(S)
to get P(W ∩ S) ?
40
40
4
4
= and P(S|W) =
= . Multiplying P(W) and P(S|W) gives
P(W|S) =
5
90
9
50
5 4
3 4
1
1
∙ = = P(W ∩ S). Multiplying P(S) and P(W|S) gives ∙ = = P(W ∩ S).
4 9
12 5
3
3
_ _
_ _ _
_ _
how to apply the formula for the probability of
independent or dependent events. Ask groups of
students to practice constructing tree diagrams for
various given independent or dependent events A
and B, and then use the multiplication rule to find
the probabilities. Have them critique each other’s
work, switch roles, and repeat the exercise several
times.
_ _ _
There are 4 green, 10 red, and 6 yellow marbles in a bag. Each time you randomly
choose a marble, you put it aside before choosing another marble at random. Use the
Multiplication Rule to find the specified probability, writing it as a fraction.
Find the probability that you choose a red marble followed by a yellow marble.
Let R be the event that the first marble is red. Let Y be the event that the second marble is
3
10 6
yellow. P(R ∩ Y) = P(R) ∙ P(Y|R) =
=
∙
19
20 19
Find the probabilty that you choose one yellow marble followed by another yellow marble.
Let Y1 be the event that the first marble is yellow. Let Y2 be the event that the second marble
5
3
6
is yellow. P(Y1 ∩ Y2) = P(Y1) ∙ P(Y2|Y1) =
∙
=
20 19
38
_ _ _
4.
_ _ _
5.
Find the probability that you choose a red marble, followed by a yellow marble, followed
by a green marble.
Let R be the event that the first marble is red. Let Y be the event that the second marble is
yellow. Let G be the event that the third marble is green.
10 6
P(R ∩ Y ∩ G) = P(R) ∙ P(Y|R) ∙ P(G|R ∩ Y) =
∙
∙ 4 = 2
57
20 19 18
Find the probability that you choose three red marbles.
_ _ _ _
6.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Julio
Etchart/Demotix/Corbis
3.
Let R1, R2, and R3 be the events that the first, second, and third marbles, respectively, are red.
10 9
8
∙
∙
= 2
P(R1 ∩ R2 ∩ R3) = P(R1) ∙ P(R2|R1) ∙ P(R3|R1 ∩ R2) =
20 19 18
19
_ _ _ _
Module 20
A2_MNLESE385900_U8M20L3 1038
1038
Lesson 3
8/28/14 8:58 PM
Dependent Events 1038
The table shows the sums that are possible when you roll two number cubes and
add the numbers. Use this information to answer the questions.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Using a smaller sample space may help
+
1
2
3
4
5
6
students see the difference between independent and
dependent events. Decrease the size of a sample space
for an exercise and have students draw probability
trees for two events. One should involve replacement
and the other should not. Have students write the
probability for each branch of the tree and use them
to find the probability for each event.
7.
1
2
3
4
5
6
2
3
4
5
6
7
3
4
5
6
7
8
4
5
6
7
8
9
5
6
7
8
9
10
6
7
8
9
10
11
7
8
9
10
11
12
Let A be the event that you roll a 2 on the number cube represented by the row
labeled 2. Let B be the event that the sum of the numbers on the cubes is 7.
a. Are these events independent or dependent? Explain.
6
1
_
= _.
6
36
1
_
Of the 6 outcomes in the row labeled 2, a sum of 7 appears once, so P(B|A) = .
Of the 36 outcomes in the entire table, a sum of 7 appears 6 times, so P(B) =
AVOID COMMON ERRORS
b. What is P(A ∩ B)?
Students should understand that P(A | B) is not the
same as P(B | A), and they should not be confused in
the formula P(A and B) = P(A) ⋅ P(B | A). Emphasize
that the order of the letters matters, and encourage
students to articulate the expressions out loud to
reinforce the difference.
8.
_ _ _
P(A ∩ B) = P(A) ⋅ P(B) = 1 · 1 = 1
6 6
36
Let A be the event that you roll a 3 on the number cube represented by the row
labeled 3. Let B be the event that the sum of the numbers on the cubes is 5.
a. Are these events independent or dependent? Explain.
1
4
_
= _.
9
36
1
_
Of the 6 outcomes in the row labeled 3, a sum of 5 appears once, so P(B|A) = .
Of the 36 outcomes in the entire table, a sum of 5 appears 4 times, so P(B) =
© Houghton Mifflin Harcourt Publishing Company
b. What is P(A ∩ B)?
9.
_ _ _
1
1
P(A ∩ B) = P(A) ∙ P(B|A) = 1 ∙ =
36
6 6
A cooler contains 6 bottles of apple juice and 8 bottles of grape juice. You choose a
bottle without looking, put it aside, and then choose another bottle without looking.
Match each situation with its probability. More than one situation can have the same
probability.
4
D
_
a. Choose apple juice and then grape juice.
13
24
A, C _
b. Choose apple juice and then apple juice.
91
15
B
_
c. Choose grape juice and then apple juice.
91
_ _ _
_ _ _
Module 20
A2_MNLESE385900_U8M20L3 1039
Lesson 20.3
6
Since P(B|A) ≠ P(B), events A and B are dependent.
d. Choose grape juice and then grape juice.
A1 = apple 1st, A2 = apple 2nd; G1 = grape 1st, G2 = grape 2nd
8
6
24
a. P(A1) ∙ P(G2|A1) =
∙
=
c. P(G1) ∙ P(A2|G1) =
14 13
91
5
15
6
∙
=
d. P(G1) ∙ P(G2|G1) =
b. P(A1) ∙ P(A2|A1) =
14 13
91
1039
6
Since P(B|A) = P(B), events A and B are independent.
1039
6
8 _
24
_
∙
=_
14 13
91
8 _
7
4
_
∙
=_
14
13
13
Lesson 3
8/26/14 12:54 PM
10. Jorge plays all tracks on a playlist with no repeats. The playlist he’s
listening to has 12 songs, 4 of which are his favorites.
a. What is the probability that the first song played is one of his
favorites, but the next two songs are not?
Let F1 be the event that the first song played is a favorite.
Let NF2 and NF3 be the events that the second and third
songs, respectively, are not favorites. Then
P(F1 ⋂ NF2 ⋂ NF3) = P(F1) ∙ P(NF2|F1) ∙ P(NF3| F1 ⋂ NF2) =
8 _
28
7
4 _
_
∙
= _.
∙
12
11
10
165
b. What is the probability that the first three songs played are all his favorites?
Let F1, F2, and F3 be the events that the first, second, and third songs, respectively, are his
3
1
2
4
∙
=
∙
.
favorites. Then P(F1 ⋂ F2 ⋂ F3) = P(F1) · P(F2|F1) · P(F3| F1 ⋂ F2) =
55
12 11 10
c. Jorge can also play the tracks on his playlist in a random order with repeats
possible. If he does this, how does your answer to part b change? Explain why.
_ _ _ _
With repeats allowed, the probability that a favorite song is played is always
4
1
___
= __
. The events F1, F2, and F3 are now independent, so you can multiply their
12
3
probabilities without considering conditional probabilities:
1
1 1 1
.
P(F1 ⋂ F2 ⋂ F3) = P(F1) · P(F2) · P(F3) = ∙ ∙ =
3 3 3
27
_ _ _ _
a. What is the probability that the next two balls drawn do not have a letter-number
combination you need, but the third ball does?
Let NC1 and NC2 be the events that the letter-number combination on the next ball and
the ball after that, respectively, are not what you need. Let C3 be the event that the letternumber combination on the third ball is what you need. Then
68
17 ___
1
P(NC1 ⋂ NC2 ⋂ C3) = P(NC1) ∙ P(NC2|NC1) ∙ P(C3|NC1 ⋂ NC2) = ___
∙ 16 ∙ __
= ____
.
20 19 3
285
b. What is the probability that none of the letter-number combinations you need is
called from the next three balls?
Let NC1, NC2, and NC3 be the events that a letter-number combination you don’t need is
called from the next ball, the ball after that, or the ball after that, respectively. Then
15
34
17 ___
P(NC1 ⋂ NC2 ⋂ NC3) = P(NC1) ∙ P(NC2|NC1) ∙ P(NC3|NC1 ⋂ NC2) = ___
∙ 16 ∙ ___
= ___
.
57
20 19 18
Module 20
A2_MNLESE385900_U8M20L3 1040
1040
© Houghton Mifflin Harcourt Publishing Company • Image Credits: (t) ©Steve
Hix/Somos Images/Corbis; (b) ©Tetra Images/Corbis
11. You are playing a game of bingo with friends. In this game,
balls are labeled with one of the letters of the word BINGO
and a number. Some of these letter-number combinations
are written on a bingo card in a 5 × 5 array, and as balls are
randomly drawn and announced, players mark their cards
if the ball’s letter-number combination appears on the cards.
The first player to complete a row, column, or diagonal on
a card says “Bingo!” and wins the game. In the game you’re
playing, there are 20 balls left. To complete a row on your
card, you need N-32 called. To complete a column, you need
G-51 called. To complete a diagonal, you need B-6 called.
Lesson 3
8/26/14 12:56 PM
Dependent Events 1040
H.O.T. Focus on Higher Order Thinking
JOURNAL
12. You are talking with 3 friends, and the conversation turns to birthdays.
Have students use mathematical notation to define
the probability of independent and dependent events.
Then have students use their own words to explain
what the notation means and how to use it.
a. What is the probability that no two people in your group were born in the
same month?
Use your birth month as a starting point. Let NB1 be the event that friend
1 has a different birth month than you. Let NB2 be the event that friend 2
has different birth month than you and friend 1. Let NB3 be the event that
friend 3 has a different birth month than you, friend 1, and friend 2. Then
55
9
11 ___
∙ 10 ∙ ___
= ___
.
P(NB1 ⋂ NB2 ⋂ NB3) = P(NB1) · P(NB2|NB1) · P(NB3|NB1 ⋂ NB2) = ___
12 12 12
96
b. Is the probability that at least two people in your group were born in the same
month greater or less than _12 ? Explain.
The event that at least two people in your group were born in the same month is the
complement of the event in part a. So, the probability that at least two people in your
55
41
1
= ___
, which is less than __
.
group were born in the same month is 1 - __
2
96
96
c.
How many people in a group would it take for the probability that at least two
people were born in the same month to be greater than _12 ? Explain.
Extend the results from part a:
55 ___
55
∙ 8 = ____
P((NB1⋂ NB2⋂ NB3)⋂ NB4) = P(NB1 ⋂ NB2 ⋂ NB3) ∙ P(NB4| NB1 ⋂ NB2 ⋂ NB3) = ___
144
96 12
The probability that at least two people in a group of 5 were born in the same month is
55
89
1
1 - ____
= ____
, which is greater than __
.
144
144
2
13. Construct Arguments Show how to extend the Multiplication Rule to three events
A, B, and C.
P(A ⋂ B ⋂ C) = P((A ⋂ B) ⋂ C)
Group events A and B as one event.
= P(A ⋂ B) ∙ P(C| A ⋂ B)
© Houghton Mifflin Harcourt Publishing Company
= P(A) ∙ P(B| A) ∙ P(C| A ⋂ B)
Apply the Multiplication Rule to A ⋂ B and C.
Apply the Multiplication Rule to A and B.
14. Make a Prediction A bag contains the same number of red marbles and blue
marbles. You choose a marble without looking, put it aside, and then choose another
marble. Is there a greater-than-50% chance or a less-than-50% chance that you choose
two marbles with different colors? Explain.
Let R1 and R2 be the events that a red marble is drawn on the first and second
draws, respectively. Let B1 and B2 be the events that a blue marble is drawn on
the first and second draws, respectively. Let n be the number of marbles of each
n ______
n
n
∙ n = _________
= ________
and
color. Then P(R1 ⋂ B2) = P(R1) ∙ P(B2|R1) = ___
2n 2n - 1
2
2n(2n - 1)
2(2n - 1)
n
n2
n
P(B1 ⋂ R2) = P(B1) ∙ P R2|B1 = 2n
∙ 2n n- 1 =
=
. Since the events
2n(2n - 1)
2(2n - 1)
(
) ___ ______ _________ ________
R1 ⋂ B2 and B1 ⋂ R2 are mutually exclusive, you can add their probabilities to get the
probability of choosing two marbles with different colors:
n
n
2n
n
1
1
________
+ ________
= ________
= ______
= ____
. Since the denominator of ____
is
1
1
2n - 1
2(2n - 1)
2(2n - 1)
2(2n - 1)
2 - __
2 - __
n
n
1
__
always less than 2 for n > 0, the fraction is always greater than . So, there is always a
2
greater-than-50% chance that you choose two marbles with different colors.
Module 20
A2_MNLESE385900_U8M20L3 1041
1041
Lesson 20.3
1041
Lesson 3
8/26/14 1:01 PM
Lesson Performance Task
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Two skydiver landing rectangles are outlined
To prepare for an accuracy landing competition, a team of skydivers has laid out targets in a
large open field. During practice sessions, team members attempt to land inside a target.
Two rectangular targets are shown on each field. Assuming a skydiver lands at random in the
field, find the probabilities that the skydiver lands inside the specified target(s).
on a 10 × 10 grid. Describe the rectangles if the given
probability is true.
1. Calculate the probabilities using the targets shown here.
a. P(A)
b. P(B)
c. P(A ⋂ B)
d. P(A ⋃ B)
e. P(A|B)
a. 0.12
• P(A and B) = 0 The rectangles do not intersect.
A
• P(A or B) = 1 The rectangles cover the entire
grid.
b. 0.2
c. 0
d. 0.32
• P(A|B) = 1 Rectangle B is contained in
rectangle A.
e. 0
f. 0
B
• P(B|A) = 0 The rectangles do not intersect.
f. P(B|A)
2. Calculate the probabilities using the targets shown here.
a. P(A)
b. P(B)
c. P(A ⋂ B)
d. P(A ⋃ B)
e. P(A|B)
a. 0.36
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 In the 10 × 10 grid for Question 2 of the
A
b. 0.14
c. 0.04
d. 0.46
2
e.
7
1
f.
9
_
_
© Houghton Mifflin Harcourt Publishing Company
f. P(B|A)
3. Calculate the probabilities using the targets shown here.
a. P(A)
a. 0.35
b. P(B)
b. 0.01
c. P(A ⋂ B)
d. 0.35
d. P(A ⋃ B)
e. 1
1
f.
35
e. P(A|B)
Lesson Performance Task, rectangle A measures 6 ×
6 units and rectangle B measures 7 × 2. Find the
dimensions and amount of overlap of two rectangles
different from rectangles A and B for which the
answers to Questions 2c and 2d are unchanged.
Sample answer: rectangle A measures 6 × 5;
rectangle B measures 10 × 2; amount of overlap:
4 squares.
B
A
c. 0.01
B
_
f. P(B|A)
Module 20
Lesson 3
1042
EXTENSION ACTIVITY
A2_MNLESE385900_U8M20L3 1042
Supply students with grid paper. Ask them to draw a 10 ×10 grid representing a
skydiver landing area. Then have them draw two rectangles, A and B,
so that P(A and B) = 0.06 and P(A or B) = 0.58. Ask them to find P(A | B) and
P(B | A). Many rectangles A and B are possible. All drawings must show two
rectangles with an overlap of 6 squares. Additionally, Area rectangle A, in its
entirety, plus Area rectangle B, in its entirety, must equal 64 square units.
Sample: rectangle A measuring 6 × 4 units overlaps by 6 squares rectangle B
3
1
measuring 8 × 5 units; P (A | B) =
and P (B | A)= .
_
20
_
4
4/2/14 2:40 PM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Dependent Events 1042