Download Conditional Probability

LESSON
20.1
Name
Conditional Probability
Class
Date
20.1 Conditional Probability
Essential Question: How do you calculate a conditional probability?
Common Core Math Standards
The student is expected to:
Explore 1
S-CP.4
Construct and interpret two-way frequency tables of data when two
categories are associated with each object being classified. Use the
two-way table as a sample space to decide if events are independent and
to approximate conditional probabilities. Also S-CP.3, S-CP.5, S-CP.6
Finding Conditional Probabilities from
a Two-Way Frequency Table
Resource
Locker
The probability that event A occurs given that event B has already occurred is called the
conditional probability of A given B and is written P(A | B).
One hundred migraine headache sufferers participated in a study of a new medicine. Some were
given the new medicine, and others were not. After one week, participants were asked if they had
experienced a headache during the week. The two-way frequency table shows the results.
Mathematical Practices
MP.4 Modeling
Took medicine
No medicine
Total
Language Objective
Headache
11
13
24
Explain to a partner how to find conditional probabilities.
No headache
54
22
76
Total
65
35
100
Let event A be the event that a participant did not get a headache. Let event B be the event that
a participant took the medicine.
ENGAGE
You can calculate the conditional probability
P(A | B) from a two-way frequency table using the
n(A ∩ B)
formula P(A | B) = _______. You can also calculate
n(B)
P(A ∩ B)
P(A | B) using the formula P(A | B) = _______.
P(B)
PREVIEW: LESSON
PERFORMANCE TASK
View the Engage section online. Discuss the
photograph. Ask students to identify the celebrity and
to describe the reasons for the celebrity’s fame. Then
preview the Lesson Performance Task.
© Houghton Mifflin Harcourt Publishing Company
Essential Question: How do you
calculate a conditional probability?
A
To the nearest percent, what is the probability that a participant who took the medicine
did not get a headache?
54
65 participants took the medicine.
So, P(A | B) = _ ≈ 83 %.
65
Of these, 54 did not get a headache.
B
To the nearest percent, what is the probability that a participant who did not get a headache
took the medicine?
54
76 participants did not get a headache.
So, P(B | A) = _ ≈ 71 %.
76
Of these, 54 took the medicine.
C
Let n(A) be the number of participants who did not get a headache, n(B) be the number of
participants who took the medicine, and n(A ⋂ B) be the number of participants who took
the medicine and did not get a headache.
n(A) = 76
n(B) = 65
n(A ∩ B) = 54
Express P(A | B) and P(B | A) in terms of n(A), n(B), and n(A ⋂ B).
n(A ⋂ B)
n(A ⋂ B)
P(A | B) = _ P(B | A) = _
n(B)
n(A)
Module 20
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1003
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© Houghto

L1 1003
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A2_MNLE
Lesson 20.1
Turn to these pages to
find this lesson in the
hardcover student
edition.

Module 20
1003
HARDCOVER PAGES 729736
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Lesson 1
8/26/14
11:01 AM
8/26/14 11:00 AM
Reflect
1.
2.
EXPLORE 1
For the question “What is the probability that a participant who did not get a headache
took the medicine?”, what event is assumed to have already occurred?
The event that a participant did not get a headache is assumed to have already occurred.
Finding Conditional Probabilities
from a Two-Way Frequency Table
In general, does it appear that P(A | B) = P(B | A)? Why or why not?
No, the calculations of P(A | B) and P(B | A) in Steps A and B show that these conditional
probabilities are not equal, so in general P(A | B) ≠ P(B | A).
Explore 2
INTEGRATE TECHNOLOGY
Finding Conditional Probabilities from
a Two-Way Relative Frequency Table
Students have the option of doing the Explore activity
either in the book or online.
You can develop a formula for P(A | B) that uses relative frequencies (which are probabilities) rather than
frequencies (which are counts).
Took medicine
No medicine
Total
Headache
11
13
24
No headache
54
22
76
Total
65
35
100
A
QUESTIONING STRATEGIES
To obtain relative frequencies, divide every number in the table by 100, the total
number of participants in the study.
Took medicine
No medicine
Total
Headache
0.11
0.13
0.24
No headache
0.54
0.22
0.76
Total
0.65
0.35
1
Recall that event A is the event that a participant did not get a headache and that event
B is the event that a participant took the medicine. Use the relative frequency table from
Step A to find P(A), P(B), and P(A ⋂ B).
P(A) = 0.76, P(B) = 0.65, and P(A ⋂ B) = 0.54.
C
In the first Explore, you found the conditional probabilities P(A | B) ≈ 83% and
P(B|A) ≈ 71% by using the frequencies in the two-way frequency table. Use the relative
frequencies from the table in Step A to find the equivalent conditional probabilities.
0.54
P(A ⋂ B)
P(A | B) = _ = _ ≈
P(B)
0.65
83 %
Module 20
What does the notation P(A | B)
represent? the conditional probability that
event A will happen given that event B has already
occurred
© Houghton Mifflin Harcourt Publishing Company
B
As you read down the column of a two-way
table, how do you find the probability that
each event happened? Sample answer: Within the
same column, you divide the number of outcomes
for each event by the total number of outcomes in
that column.
EXPLORE 2
Finding Conditional Probabilities
from a Two-Way Relative Frequency
Table
0.54
P(A ⋂ B)
P(B | A) = _ = _ ≈ 71 %
P(A)
0.76
1004
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Two-way frequency tables are efficient ways to
Lesson 1
PROFESSIONAL DEVELOPMENT
A2_MNLESE385900_U8M20L1 1004
Learning Progressions
This is the first of three related lessons that cover the concepts of conditional
probability, independent events, and dependent events. Many texts begin by
defining independent and dependent events. The approach in this module begins
with a deeper treatment of conditional probability and then progresses to defining
independent and dependent events. This approach gives students many
opportunities to work with two-way tables, which they can use to make sense of a
wide range of probability and statistics problems as they continue their study of
mathematics.
8/26/14 11:02 AM
express quantitative data that can be categorized by
two variables. Point out that there are two formulas for
n(A ∩ B)
P(A ∩ B)
P(A | B): P(A | B) = ______ and P(A | B) = _______ .
n(B)
P(B)
It is important for students to be able to identify each
of the quantities in the two formulas, and for them to
distinguish between n(B), the number of outcomes
for B, and P(B), the probability of B.
Conditional Probability
1004
Generalize the results by using n(S) as the number of elements in the sample space

QUESTIONING STRATEGIES
(in this case, the number of participants in the study). For instance, you can write
n(A)
P(A) = ____. Write each of the following probabilities in a similar way.
n(S)
(A ⋂ B)
n
_
What does A ∩ B represent in a two-way
table? the number of outcomes that
represent both events, A and B
What does n(S) represent? the total number
of outcomes in the sample space
n(S)
P(A ⋂ B)
P(A | B) = _ = _
(
)
B
n
_
P(B)
n(S)
Reflect
n(A ⋂ B)
P(A ⋂ B)
Why are the two forms of P(A ⋂ B), _______ and _______, equivalent?
n(B)
P(B)
(A ⋂ B) ÷ n(S)
)
(
n___________
P
A
⋂
B
= ______
n(B) ÷ n(S)
P(B)
3.
EXPLAIN 1
What is a formula for P(B | A) that involves probabilities rather than counts? How do you
4.
_______
Explain 1
AVOID COMMON ERRORS
n(A ⋂ B)
obtain this formula from the fact that P(B | A) = _______?
n(A)
P(A ⋂ B)
n(A ⋂ B)
; you divide the numerator and denominator of
by n(S).
P(B | A) =
P(A)
n(A)
Using the Conditional Probability
Formula
_______
Using the Conditional Probability Formula
In the previous Explore, you discovered the following formula for conditional probability.
Conditional Probability
The conditional probability of A given B (that is, the probability that event A occurs given
that event B occurs) is as follows:
P(A ⋂ B)
P(A | B) = _
P(B)
Example 1
© Houghton Mifflin Harcourt Publishing Company
Students may be confused about how to find the total
number of outcomes in a two-way table. Tell them
that the total number of outcomes (the number of
outcomes in the sample space) should appear at the
lower right corner. It equals the total of the bottom
row of the table and the total of the right-most
column of the table.
n(A ⋂ B)
P(A ⋂ B) = _
n(S)
n(B)
P(B) = _
n(S)

Find the specified probability.
For a standard deck of playing cards, find the probability that a
red card randomly drawn from the deck is a jack.
Step 1 Find P(R), the probability that a red card is drawn from the deck.
26 .
There are 26 red cards in the deck of 52 cards, so P(R) = _
52
Step 2 Find P(J ⋂ R), the probability that a red jack is drawn from the deck.
2.
There are 2 red jacks in the deck, so P(J ⋂ R) = _
52
Module 20
1005
Lesson 1
COLLABORATIVE LEARNING
A2_MNLESE385900_U8M20L1 1005
Small Group Activity
Give groups of students sample two-way tables. Have one student verify all of the
totals in the table and give the size of the sample. Ask a second student to explain
how to find the conditional probabilities of one column of the table. Then have a
third student explain how to find the conditional probabilities of the second
column. Have a fourth student verify that the conditional probabilities for the
table are consistent with the sample space and are correct. Then have students
compare their results, and present them to the class.
1005
Lesson 20.1
8/26/14 11:14 AM
B
Step 3 Substitute the probabilities from Steps 1 and 2 into the formula for conditional
probability.
2
_
P(J ⋂ R)
52
P(J | R) = _ = _
26
_
P(R)
52
Step 4 Simplify the result.
2 ⋅ 52
_
52
1
2 =_
P(J | R) = _
=_
13
26
26 ⋅ 52
_
52
For a standard deck of playing cards, find the probability that a jack randomly drawn
from the deck is a red card.
QUESTIONING STRATEGIES
What is the conditional probability formula,
and what does it represent?
P(A | B) = _______; the probability that event A
P(B)
occurs given that event B has already occurred is
represented by P(A | B), and is equal to the
probability that both A and B occur divided by the
probability that event B occurs.
P(A ∩ B)
Step 1 Find P(J), the probability that a jack is drawn from the deck.
4
There are 4 jacks in the deck of 52 cards, so P(J) = _.
52
Step 2 Find P(J ⋂ R), the probability that a red jack is drawn from the deck.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Have students follow a pattern as they find
2
There are 2 red jacks in the deck, so P(J ⋂ R) = _.
52
Step 3 Substitute the probabilities from Steps 1 and 2 into the formula for conditional
probability.
2
_
P(J ⋂ R)
52
_
_
P(J | R) =
=
P(R)
4
_
52
Step 4 Simplify the result.
the values to substitute into the conditional
P(A ∩ B)
probability formula P(A | B) = ______. First,
P(B)
identify which event has already occurred (B, in this
case) and find the probability that B occurs (divide
the number of B outcomes by the number in the
sample space). Then find the number of events that
are in the intersection of A and B and calculate the
probability that both A and B occur (the sum of A
and B divided by the number in the sample space).
Finally, substitute the values and divide.
2
_ ⋅ 52
2
52
1
_
P(J | R) =
= _= _
4
4
2
_ ⋅ 52
52
© Houghton Mifflin Harcourt Publishing Company
Your Turn
5.
For a standard deck of playing cards, find the probability that a face card randomly
drawn from the deck is a king. (The ace is not a face card.)
Let F be the event of drawing a face card and K be the event of drawing a king. Then
4
P(F ⋂ K)
52
12
4
= 4 = 1.
=
and P(F ⋂ K) =
, so P(K | F) =
P(F) =
52
12
52
12
3
P(F)
52
For a standard deck of playing cards, find the probability that a queen randomly drawn
from the deck is a diamond.
_
6.
_
_ _ _ _
_
_
Let Q be the event of drawing a queen and D be the event of drawing a diamond. Then
1
P(Q ⋂ D)
52
4
1
= 1.
and P(Q ⋂ D) =
P(Q) =
, so P(D|Q) =
=
4
52
52
4
P(Q)
52
_
_
_ _ _
_
_
Module 20
1006
Lesson 1
DIFFERENTIATE INSTRUCTION
A2_MNLESE385900_U8M20L1 1006
8/27/14 4:53 PM
Modeling
Some students may benefit from a hands-on approach for finding the conditional
probabilities for a two-way table. Have groups of students complete a list of
questions that they can ask about the table below, such as, “What is the
probability that a household owns a dog given that the household owns a
cat?” 15 ÷ 33 ≈ 0.45
Owns a cat
Owns a
dog
Yes
No
Yes
15
24
No
18
43
Conditional Probability
1006
Elaborate
ELABORATE
7.
QUESTIONING STRATEGIES
When calculating a conditional probability from a two-way table, explain why it doesn’t
matter whether the table gives frequencies or relative frequencies.
A conditional probability is a ratio of two frequencies. If you divide those frequencies by
the same number to convert them to relative frequencies, their ratio remains unchanged.
How can you find conditional probabilities for
the data in a two-way table? Label the events
represented by the table as A and B and use the
n(A ∩ B)
formula P(A | B) = _______ .
8.
Discussion Is it possible to have P(B | A) = P(A | B) for some events A and B? What
conditions would need to exist?
Yes, it is possible for P(B | A) to equal P(A | B). This would happen if the probability of event
A is equal to the probability of event B.
n(B)
9.
SUMMARIZE THE LESSON
Given a two-way frequency table, how could
you quickly verify that it has been filled out
correctly? How could you do the same for a two-way
relative frequency table? You could check that the
sum of the frequencies in the bottom row and the
sum of the frequencies in the right-most column are
equal to each other and to the number of outcomes
in the sample space; in a two-way relative frequency
table, both sums should be equal to 1.
Essential Question Check-In In a two-way frequency table, suppose event A represents
a row of the table and event B represents a column of the table. Describe how to find the
conditional probability P(A | B) using the frequencies in the table.
Divide the frequency that appears in the intersection of the row for A and the column for B
by the total of all frequencies in the row for B.
Evaluate: Homework and Practice
© Houghton Mifflin Harcourt Publishing Company
In order to study the relationship between the amount of sleep a student gets and
his or her school performance, a researcher collected data from 120 students. The
two-way frequency table shows the number of students who passed and failed an
exam and the number of students who got more or less than 6 hours of sleep the
night before. Use the table to answer the questions in Exercises 1–3.
1.
Passed exam
Failed exam
Total
Less than 6 hours of sleep
12
10
22
More than 6 hours of sleep
90
8
98
Total
102
18
120
• Online Homework
• Hints and Help
• Extra Practice
To the nearest percent, what is the probability that a student who failed the exam got
less than 6 hours of sleep?
Let L be the event of getting less than 6 hours of sleep and F be the event
n(L ⋂ F)
10
5
=
= ≈ 56%.
of failing the exam. Then P(L | F) =
18
9
n(F)
_ _ _
2.
To the nearest percent, what is the probability that a student who got less than 6 hours
of sleep failed the exam?
Let L be the event of getting less than 6 hours of sleep and F be the event
n(L ⋂ F)
5
10
=
≈ 45%.
of failing the exam. Then P(F | L) =
=
11
22
n(L)
_ _ _
Module 20
1007
Lesson 1
LANGUAGE SUPPORT
A2_MNLESE385900_U8M20L1 1007
Connect Vocabulary
To help students describe the events represented in a two-way table and then
determine the numbers used to find the conditional probabilities, have them first
draw a line around the row or column of a two-way table representing the given
information. Then ask them to circle the cell for the probability they are finding in
another color. Then ask them to divide the number in that cell by the total in the
respective row or column to find the conditional probability.
1007
Lesson 20.1
8/26/14 11:14 AM
3.
To the nearest percent, what is the probability that a student got less than 6 hours of
sleep and failed the exam?
EVALUATE
Let L be the event of getting less than 6 hours of sleep, F be the event
n(L ⋂ F)
=
of failing the exam, and S be the sample space. P(L ⋂ F) =
n(S)
10
≈ 8%.
120
_
_
4.
You have a standard deck of playing cards from which you randomly select a card.
Event D is getting a diamond, and event F is getting a face card (a jack, queen, or
king).
n(D ⋂ F)
P(D ⋂ F)
Show that P(D | F) = _ and P(D | F) = _ are equal.
n(F)
P(F)
3
3 ⋅ 52
_
n(D ⋂ F)
P(D ⋂ F)
52
3
3
1
52
=_
=
P(D | F) =
=
= 1
=
=
P(D | F) =
12 ⋅ 52
4
4
12
12
12
_
n(F)
P(F)
52
52
_
_ _
_
_ _ _
ASSIGNMENT GUIDE
_ _
The table shows data in the previous table as relative frequencies (rounded to the
nearest thousandth when necessary). Use the table for Exercises 5–7.
5.
Passed exam
Failed exam
Total
Less than 6 hours of sleep
0.100
0.083
0.183
More than 6 hours of sleep
0.750
0.067
0.817
Total
0.850
0.150
1.000
To the nearest percent, what is the probability that a student who passed the exam got
more than 6 hours of sleep?
Let M be the event of getting more than 6 hours of sleep and Pa be the event
P(M ⋂ Pa)
0.750
of passing the exam. Then P(M | Pa) =
≈ 0.882 ≈ 88%.
=
0.850
P(Pa)
_ _
To the nearest percent, what is the probability that a student who got more than
6 hours of sleep passed the exam?
Let M be the event of getting more than 6 hours of sleep and Pa be the event of
n(M ⋂ Pa)
0.750
passing the exam. Then P(Pa | M) =
= 0.918 ≈ 92%.
=
0.817
n(M)
_ _
7.
Which is greater, the probability that a student who got less than 6 hours of sleep
passed the exam or the probability that a student who got more than 6 hours of sleep
failed the exam? Explain.
Let L be the event of getting less than 6 hours of sleep, and let M be the event of getting
Module 20
Exercise
A2_MNLESE385900_U8M20L1 1008
_ _
Depth of Knowledge (D.O.K.)
Mathematical Practices
MP.5 Using Tools
2 Skills/Concepts
MP.4 Modeling
1 Recall of Information
MP.5 Using Tools
14
2 Skills/Concepts
MP.4 Modeling
15
2 Skills/Concepts
MP.2 Reasoning
16–20
7
8–13
Exercises 1–3,
14–15, 20, 22–23
Explore 2
Finding Conditional Probabilities
from a Two-Way Relative Frequency
Table
Exercises 4–7,
18–19
Example 1
Using the Conditional Probability
Formula
Exercises 9–13,
16–17, 21
Lesson 1
1008
1 Recall of Information
1–6
Explore 1
Finding Conditional Probabilities
from a Two-Way Frequency Table
quantitative data that can be categorized by two
variables. Joint relative frequencies are the values in
each category divided by the total number of values,
while marginal relative frequencies are found by
adding the joint relative frequencies in each row or
column. A conditional relative frequency is the
quotient of a joint relative frequency and the
marginal relative frequency. Conditional relative
frequencies give an alternate way to find conditional
probabilities.
more than 6 hours of sleep. Let Pa be the event of passing the exam, and let F be the event
P(L ⋂ Pa)
0.100
=
≈ 0.546 ≈ 55%, and
of failing the exam. Then P(Pa | L) =
0.183
P(L)
P(M ⋂ F)
0.067
P(F | M) =
=
≈ 0.082 ≈ 8%, so the probability that a student who got less
0.817
P(M)
than 6 hours of sleep passed the exam is greater.
_ _
Practice
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Two-way frequency tables express
© Houghton Mifflin Harcourt Publishing Company
6.
Concepts and Skills
2 Skills/Concepts
MP.5 Using Tools
21
3 Strategic Thinking
MP.6 Precision
22
3 Strategic Thinking
MP.3 Logic
23
3 Strategic Thinking
MP.2 Reasoning
8/26/14 11:14 AM
Conditional Probability
1008
You randomly draw a card from a standard deck of playing cards. Let A be the
event that the card is an ace, let B be the event that the card is black, and let C be
the event that the card is a club. Find the specified probability as a fraction.
AVOID COMMON ERRORS
Students may assume that B always represents the
event assumed to have taken place. Explain that A
and B are simply variables and that the conditional
probability formula holds true no matter which
letters are used. For example, the formula can be used
to find the probability of B given A:
8.
P(A | B) P(A | B) =
P(A ⋂ B) _
_
= 1
P(A ⋂ B) _
_
= 1
P(B)
P(A)
13
10. P(A | C) P(A | C) =
11. P(C | A)
P(C)
P(A ∩ B)
P(A)
P(C | A) =
P(A)
13
12. P(B | C) P(B | C) =
the vertical bar represents the event assumed to have
taken place, and that its probability will always be in
the denominator of the fraction.
2
P(A ⋂ C) _
_
= 1
P(A ⋂ C) _
_
= 1
P(B | A) = ______. Emphasize that the letter after
P(B | A) P(B | A) =
9.
13. P(C | B)
4
P(C | B) =
P(B ⋂ C) _
_
= 1
P(B ⋂ C)
_
=1
P(B)
P(C)
2
14. A botanist studied the effect of a new fertilizer by choosing 100 orchids and giving
70% of these plants the fertilizer. Of the plants that got the fertilizer, 40% produced
flowers within a month. Of the plants that did not get the fertilizer, 10% produced
flowers within a month.
© Houghton Mifflin Harcourt Publishing Company • ©Design Pics/Superstock
a. Use the given information to complete the two-way frequency table.
Received
fertilizer
Did not receive
fertilizer
Total
Did not flower
in one month
70 - 28 = 42
30 - 3 = 27
42 + 27 = 69
Flowered in one
month
0.4 ∙ 70 = 28
0.1 ∙ 30 = 3
28 + 3 = 31
Total
0.7 ∙ 100 = 70
0.3 ∙ 100 = 30
100
b. To the nearest percent, what is the probability that an orchid that produced
flowers got fertilizer?
Let Fl be the event that an orchid produced flowers and Fe be the event that an
n(Fe ⋂ Fl)
28
orchid got fertilizer. Then P(Fe | FI) =
≈ 90%.
=
31
n(FI)
c. To the nearest percent, what is the probability that an orchid that got fertilizer
produced flowers?
_ _
Then P(Fl | Fe) =
Module 20
A2_MNLESE385900_U8M20L1 1009
1009
Lesson 20.1
n(Fe ⋂ Fl) _
28 _
_
= 2 = 40%.
=
n(Fe)
70
5
1009
Lesson 1
4/3/14 2:00 PM
15. At a school fair, a box contains 24 yellow balls and 76 red balls. One-fourth of the
balls of each color are labeled “Win a prize.” Match each description of a probability
with its value as a percent.
B 76%
A. The probability that a randomly selected
ball labeled “Win a prize” is yellow
D 25%
B. The probability that a randomly selected
ball labeled “Win a prize” is red
C 19%
D. The probability that a randomly selected
yellow ball is labeled “Win a prize”
Winner
Not winner
Total
Red
_1 (76) = 19
_3 (76) = 57
76
Yellow
_1 (24) = 6
_3 (24) = 18
24
25
75
100
4
Total
interpret the tables in terms of percentages. For
example, if 10 students in the class own a cat, and 6 of
those students also own a dog, have students
verbalize that the conditional probability of owning a
6
= 60%.
dog given that they own a cat is __
10
A 24%
C. The probability that a randomly selected
ball is labeled “Win a prize” and is red
4
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 After making a two-way table, have students
4
4
Let R be the even that a ball is red, Y be the event that a ball is yellow, W be the event that a
ball is labeled “Win a prize,” and S is the sample space.
A. P(Y | W) =
n(Y ⋂ W) _
6
_
= 24%
=
n(W)
n(R ⋂ W)
25
19
B. P(R | W) = _ = _ = 76%
n(W)
25
C. P(W ∩ R) =
n(W ⋂ R) _
19
_
= 19%
=
n(S)
100
n (Y ⋂ W )
6
D. P(W | Y) = _ = _ = 25%
n(Y)
24
16. A teacher gave her students two tests. If 45% of the students passed both tests and
60% passed the first test, what is the probability that a student who passed the first
test also passed the second?
Let T 1 be the event that a student passed the first test and T 2 be the event that the student
© Houghton Mifflin Harcourt Publishing Company
passed the second test.
P(T 1 ⋂ T 2)
0.45
= 0.75 = 75%
P(T 2 | T 1) = _ =
0.60
P(T 1)
17. You randomly select two marbles, one at a time, from a pouch containing blue and
green marbles. The probability of selecting a blue marble on the first draw and a
green marble on the second draw is 25%, and the probability of selecting a blue
marble on the first draw is 56%. To the nearest percent, what is the probability of
selecting a green marble on the second draw, given that the first marble was blue?
_
Let B be the event of selecting a blue marble on the first draw and G be the event of
selecting a green marble on the second draw.
P(B ⋂ G)
0.25
=
≈ 0.45 = 45%
P(G | B) =
0.56
P(B)
_ _
Module 20
A2_MNLESE385900_U8M20L1 1010
1010
Lesson 1
4/3/14 2:00 PM
Conditional Probability
1010
You roll two number cubes, one red and one blue. The table shows the probabilities
for events based on whether or not a 1 is rolled on each number cube. Use the
table to find the specified conditional probability, expressed as a fraction. Then
show that the conditional probability is correct by listing the possible outcomes
as ordered pairs of the form (number on red cube, number on blue cube) and
identifying the successful outcomes.
Rolling a 1 on the
red cube
Not rolling a 1 on
the red cube
Total
Rolling a 1 on the
blue cube
1
_
36
5
_
36
1
_
6
Not rolling a 1 on
the blue cube
5
_
36
25
_
36
5
_
6
Total
1
_
6
5
_
6
1
18. P(not rolling a 1 on the blue cube | rolling a 1 on the red cube)
Let N1B be the event that a 1 is not rolled on the blue cube and 1R be the event that a 1 is
rolled on the red cube.
P(N1B | 1R) =
___5
__1 ∙ 36
___5
__1
∙ 36
n(N1B ∩ 1R) _
5
36
36
__
=
=_=_
n(1R)
6
6
6
Given that rolling a 1 on the red number cube has occurred, there are 6 possible outcomes
(where the ordered pairs give the number on the red cube first): (1, 1), (1, 2), (1, 3), (1, 4),
(1, 5), and (1, 6). Of these, the last 5 outcomes are successful because they involve a number
that is not 1 on the blue cube. So, the probability of not rolling a 1 on the blue cube when a 1
5
is rolled on the red cube is .
6
19. P(not rolling a 1 on the blue cube | not rolling a 1 on the red cube)
© Houghton Mifflin Harcourt Publishing Company
_
Let N1B be the event that a 1 is not rolled on the blue cube and N1R be the event that a 1 is
not rolled on the red cube.
P(N1B | N1R) =
25
___
__5
25
___
__5 ∙ 36
∙ 36
n(N1B ∩ N1R) _
5
25
36
36
__
=
=_=_=_
n(N1R)
6
6
30
6
Given that not rolling a 1 on the red number cube has occurred, there are 30 possible
outcomes (where the ordered pairs give the number on the red cube first): (2, 1), (2, 2),
(2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4)
, (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), and
(6, 6). Of these, all but 5 outcomes—(2, 1), (3, 1), (4, 1,) (5, 1), and (6, 1)—are successful
because they involve a number that is not 1 on the blue cube. So, the probability of not
5
25
= .
rolling a 1 on the blue cube when a 1 is not rolled on the red cube is
6
36
_ _
Module 20
A2_MNLESE385900_U8M20L1 1011
1011
Lesson 20.1
1011
Lesson 1
8/26/14 11:14 AM
20. The table shows the results of a quality-control study at a computer
factory.
Defective
Not defective
Total
Shipped
Not shipped
3
7
Total
10
89
1
90
92
8
100
a. To the nearest tenth of a percent, what is the probability that a
shipped computer is not defective?
b. To the nearest tenth of a percent, what is the probability that a
defective computer is shipped?
Let S be the event that a computer is shipped, D be the event that a computer is defective,
and Nd be the even that a computer is not defective.
n(Nd ⋂ S)
89
a. P(Nd | S) =
=
≈ 0.967 = 96.7%
92
n(S)
_ _
b. P(S | D) =
n(D ⋂ S) _
3
_
=
= 0.3 = 30%
n(D)
10
H.O.T. Focus on Higher Order Thinking
Since P(A | B) =
A
S
B
n(A ∩ B)
_
, you can think of P(A | B) as the ratio of the area of the overlap
n(B)
of the circles to the area of the entire circle labeled B. Since more of circle B lies inside the
n(A ∩ B)
1
,
overlap than outside the overlap, P(A | B) > . On the other hand, since P(B | A) =
2
n(A)
you can think of P(B | A) as the ratio of the area of the overlap of the circles to the area of the
_
_
entire circle labeled A. Since less of circle A lies inside the overlap than outside the overlap,
1
1
P(B | A) < . So, P(A | B) > > P(B | A).
2
2
_
Module 20
A2_MNLESE385900_U8M20L1 1012
_
1012
© Houghton Mifflin Harcourt Publishing Company • ©Ryan Pyle/Corbis
21. Analyze Relationships In the Venn diagram, the circles
representing events A and B divide the sample space S into four
regions: the overlap of the circles, the part of A not in the overlap, the
part of B not in the overlap, and the part of S not in A or B. Suppose
that the area of each region is proportional to the number of outcomes
that fall within the region. Which conditional probability is greater:
P(A | B) or P(B | A)? Explain.
Lesson 1
4/3/14 2:00 PM
Conditional Probability
1012
22. Explain the Error A student was asked to use the table shown to find the
probability, to nearest percent, that a participant in a study of a new medicine for
migraine headaches did not take the medicine, given that the participant reported no
headaches.
PEERTOPEER DISCUSSION
Give students each an example of a two-way table.
Ask them to discuss with a partner how to interpret
the two variables represented by the table. Then have
them describe in their own words the process they
would use to find the conditional probabilities for the
data given in each cell of the two-way tables.
Took medicine
No medicine
Total
Headache
11
13
24
No headache
54
22
76
Total
65
35
100
The student made the following calculation.
22 ≈ 0.63 = 63%
P(no medicine | no headache) = _
35
Explain the student’s error, and find the correct probability.
The student divided 22, the number of participants who did not take the medicine
JOURNAL
Have students explain conditional probability in their
own words. Also, ask students to give an example of
how to use the formula for conditional probability.
and did not report headaches, by 35, the number of participants who did not take
the medicine. However, the student should have divided 22 by 76, the number
of participants who reported no headaches. The correct calculation is
22
P(no medicine | no headache) =
≈ 0.29= 29%.
76
_
23. Communicate Mathematical Ideas Explain how a conditional probability
based on a two-way frequency table effectively reduces it to a one-way table. In your
explanation, refer to the two-way table shown, which lists frequencies for events A, B,
and their complements. Highlight the part of the table that supports your explanation.
A
Not A
Total
n(A ∩ B)
n(not A ∩ B)
n(B)
Not B
n(A ∩ not B)
n(not A ∩ not B)
n(not B)
Total
n(A)
n(not A)
n(S)
© Houghton Mifflin Harcourt Publishing Company
B
The conditional probability P(A | B) =
n(B)
that event is assumed to have occurred. The numbers used to calculate P(A | B) both come
from the highlighted row in the table: n(A ∩ B) is the number of outcomes in event B that
are also in event A, while n(B) is the number of all outcomes in event B. The rest of the table
is irrelevant.
Module 20
A2_MNLESE385900_U8M20L1 1013
1013
Lesson 20.1
n(A ∩ B)
_
restricts the discussion to event B because
1013
Lesson 1
4/3/14 2:00 PM
Lesson Performance Task
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 When dealing with tables like the ones in the
The two-way frequency table gives the results of a survey that asked students this question:
Which of these would you most like to meet: a famous singer, a movie star, or a sports star?
Famous singer
Movie star
Sports star
Total
20
15
55
90
Boys
Girls
40
50
20
110
Total
60
65
75
200
Lesson Performance Task and the Extension Activity,
students can check that the cell entries are consistent
by (a) adding the row entries (for the Extension
Activity, Boys + Girls = 155 + 145 = 300);
(b) adding the column entries (Famous Singers +
Movie Stars + Sports Star = 80 + 90 + 130 = 300);
and (c) checking to see that the sums are equal
(300 = 300).
a. Complete the table by finding the row totals, column totals,
and grand total.
b. To the nearest percent, what is the probability that a student
who chose “movie star” is a girl?
c. To the nearest percent, what is the probability that a student
who chose “famous singer” is a boy?
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Use the numbers in the “Movie Star” column
d. To the nearest percent, what is the probability that a boy chose
“sports star”?
e. To the nearest percent, what is the probability that a girl chose
“famous singer”?
f. To the nearest percent, what is the probability that a student
who chose either “famous singer” or “movie star” is a boy?
of the Lesson Performance Task (15 and 50). Write at
least four questions beginning with “Find the
probability of___” that can be answered by writing a
fraction that has either 15 or 50 in the numerator. For
each question, give the fraction. Sample answers:
Find the probability that a student who chose
15
“Movie Star” is a boy ___
; find the probability that a
65
50
student who chose “Movie Star” is a girl ___
; find
65
the probability that a student who is a boy chose
15
“Movie Star” ___
; find the probability that a student
90
50
who is a girl chose “Movie Star” ___
.
110
g. To the nearest percent, what is the probability that a girl did
not choose “sports star”?
_ _
c. P(B | FS) =
n(B ∩ FS) _
20
_
=
≈ 33%
d. P(SS | B) =
n(B ∩ SS) _
55
_
=
≈ 61%
e. P(FS | G) =
n(FS)
60
n(B)
90
n(G ∩ FS)
_ _
40
=
≈ 36%
110
n(G)
n(B ∩ (FS ∪ MS))
35
=
≈ 28%
f. P(B |(FS ∪ MS)) =
125
n(FS ∪ MS)
© Houghton Mifflin Harcourt Publishing Company
For the following answers, let B be the event that a student is a boy, let G be the event that a
student is a girl, let FS be the event that a student chose “famous singer,” let MS be the event that
a student chose “movie star,” and let SS be the event that a student chose “sports star.”
n(G ∩ MS)
50
≈ 77%
=
b. P(G | MS) =
65
n(MS)
( )
( )
( )
( )
__ _
g. P(not SS | G) =
n(G ∩ not SS) _
90
__
=
≈ 72%
n(not SS)
125
Module 20
Lesson 1
1014
EXTENSION ACTIVITY
A2_MNLESE385900_U8M20L1 1014
Have students fill out a table like the one in the Lesson Performance Task, using
these facts: 300 students took part in the survey described in the Task; P(student
7 ; P(student who chose “Sports Star” is a
who chose “Famous Singer” is a girl) = ___
10
4 ; of those who chose “Movie Star,” 30% are boys; 80 students chose
boy) = __
5
“Famous Singer”; 130 students chose “Sports Star.”
Famous Singer
Movie Star
Sports Star
Boys
24
27
104
Girls
56
63
26
8/26/14 11:30 AM
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.
Conditional Probability
1014