Download Alg II CC-15 TE Conditional Probability

CC-15
1 ​Interactive Learning
Conditional Probability
Solve It!
Common Core State Standards
MACC.912.S-CP.2.6 Find the conditional probability
of A given B as the fraction of B’s outcomes that
also belong to A . . . Also MACC.912.S-CP.1.1,
MACC.912.S-CP.1.3, MACC.912.S-CP.1.4,
MACC.912.S-CP.1.5, MACC.912.S-CP.2.8
MP 1, MP 2, MP 3, MP 4, MP 6
PURPOSE ​To analyze a conditional probability
Objectives To find conditional probabilities
To use tables and tree diagrams to determine conditional probabilities
situation
PROCESS ​Students may
•act out the situation.
•draw a branching tree diagram.
FACILITATE
Q What is the probability that you originally guessed
the correct box? In this case, what wins the prize? ​
[ 13 ; not switching boxes]
This sounds like
a great idea for
a television game
show!
Q What is the probability that you did not guess the
correct box? In this case, what wins the prize? ​
[ 23; switching boxes]
Q How does the probability of winning by switching
compare to the probability of winning by not
switching? ​[The probability of winning by
A great prize is inside Box 1, 2, or 3.
You choose Box 1. You are shown
that Box 2 contains a rubber chicken.
You are given a chance to change your
choice. Should you stay with Box 1 or
should you change to Box 3? Justify
your answer.
MATHEMATICAL
Lesson
Vocabulary
•conditional
probability
•contingency
table
ANSWER ​See Solve It in Answers on next page.
CONNECT THE MATH ​In the Solve It, students
Essential Understanding Conditional probability exists when two events
You write the conditional probability of event B, given that event A occurs, as P (B A).
You read P (B A) as “the probability of event B, given event A.”
A contingency table, or two-way frequency table, is a frequency table that contains data
from two different categories. Contingency tables and tree diagrams can help you find
conditional probabilities.
Problem 1 Finding Conditional Probability
A What is P (female graduate school)?
Problem 1
What’s the condition?
The student is at a
graduate school.
Q Multiply the probabilities of selecting a female and
selecting a graduate student. Is the product the
same as the probability in 1A? Explain. ​[No; it is
The condition that the person selected
is at graduate school limits the sample
space to the 3,303,000 graduate students.
Of those, 1,954,000 are female.
1954
P (female graduate school) = 3303 ≈ 0.59
54
Chapter 11
54
A conditional probability is the probability that one
event occurs, given that another event has occurred.
Math Background
Conditional probability is the probability that one
event will occur given that another event has
occurred. For two events A and B, the conditional
probability that B will occur, given that A has
occurred, is written P (B 0 A).
Once it is known that an event A has occurred, the
sample space is restricted to the subset of events
that include A. For example, if you are rolling a
number cube, then the original sample space is
51, 2, 3, 4, 5, 66. If A is the event that an even
number is rolled, and you know that A occurs, then
the new sample space is 52, 4, 66. If B is the event
that a 2 or 4 is rolled, then P (B) = 13. However,
P (B 0 A) = 23.
54 Common Core
Males
(in thousands)
Females
(in thousands)
Two-year
colleges
1866
2462
Four-year
colleges
4324
5517
Graduate
schools
1349
1954
SOURCE: U.S. Census Bureau
Probability and Statistics
HSM15_A2Reg_SE_CC_15_TrKit.indd
54
Common Core
Essential Understanding
Student Genders
Education The table shows students by
gender and by type of school in 2005. You
pick a student at random.
2 ​Guided Instruction
Big idea Probability
What’s th
The waste
be recycled
are dependent.
calculate conditional probability based on an initial
choice. In the lesson, students will find conditional
probability of situations using formulas and tree
diagrams.
CC-15 Preparing to Teach
2 3
PRACTICES The probability that an event, B, will occur given that another event, A, has already
occurred is called a conditional probability.
switching is twice that of not switching.]
much lower, about 0.11. Selecting a grad student
is a given in 1A, so that part of the probability is
100%.]
1
Recall that if two events are
independent,
then the occurrence of
HSM15_A2Reg_SE_TrKit.indd Page 54 03/08/13 3:53 PM gg-018
one event does not affect the probability
that the second event will occur. Thus,
if C and D are independent events, then
P (D 0 C ) = P (D).
Mathematical Practices
Attend to precision. Students will use a
clear definition of the term “conditional
probability” and determine when to
calculate it.
HSM15_A2Reg_SE_CC_15_TrKit.ind
02/08/13 3:04 PM
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/A ...
Problem 1
B What is P (female)?
ty
2462 + 5517 + 1954
total number of females
Q In 1B, is 0.57 a reasonable answer? Explain. ​[Yes;
P (F) = total number of students = 1866 + 2462 + 4324 + 5517 + 1349 + 1954
the number of females is greater than the
number of males in every row of the table, so
the probability of picking a female must be
greater than 50%.]
9933
= 17,472 ≈ 0.57
Got It? 1. a. In Problem 1, what is P (Four@year male)?
b. Reasoning Without calculating, given a student is enrolled in a four-year
college, is it more likely for the student to be male or female? Explain.
Got It?
Problem 2 Conditional Probability in Statistics
What’s the condition?
The waste sample has to
be recycled waste.
Q How would you translate 1a into a question using
only words? ​[What is the probability that a
Municipal Waste Collected
(millions of tons)
Multiple Choice Americans recycle
increasing amounts through municipal
waste collection. The table shows the
collection data for 2007. What is the
probability that a sample of recycled
waste is paper?
Material
Recycled
Paper
45.2
37.8
Metal
7.2
13.6
3.2
10.4
16%
33%
Glass
28%
57%
Plastic
2.1
28.6
Other
21.7
46.3
The given condition is that the waste is
recycled. A favorable outcome is that
the recycled waste is paper.
student attends a four-year college, given that
the student is male?]
Not Recycled
Problem 2
Q Based on the answer, how would you find the
probability that a sample of recycled waste is
not paper without calculating? ​[The probability
SOURCE: U.S. Environmental Protection Agency
of recycled waste that is not paper is the
probability of all the recycled waste minus the
probability of recycled waste that is paper:
100% − 57% = 43%.]
45.2
P (paper recycled) = 45.2 + 7.2 + 3.2 + 2.1 + 21.7
≈ 0.57, or 57%
The probability that the recycled waste is paper is about 57%. The correct answer is D.
Got It?
Got It? 2. a. What is the probability that a sample of recycled waste is plastic?
b. What is the probability that a sample of recycled waste is glass?
Q Does the question “What is the probability that a
sample of waste is recycled glass?” differ from the
question in 2b? Explain. ​[Yes; in 2b, it is a given
You can use a formula to find conditional probability.
that the sample is recycled.]
s)
Key Concept Conditional Probability
Take Note
For any two events A and B with P (A) ≠ 0,
P(A and B)
P(B A) =
P (A)
Q In the formula for conditional probability, is A or B
the given condition? ​[A is the given condition.]
Q Why must P (A) not equal zero? ​[Division by zero
is not allowed.]
Lesson 11-4
HSM15_A2Reg_SE_CC_15_TrKit.indd
Page 55 02/08/13 9:12 PM gg-018
02/08/13 3:04 PM
Conditional Probability
55
CC-15 Conditional
Probability
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/A
...
55
Answers
HSM15_A2Reg_SE_TrKit.indd Page 55 03/08/13 3:53 PM gg-018
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/A ...
Solve It!
Change to box 3; there is a 13 chance that box 1 has
the prize and a 23 chance that box 3 has the prize.
Got It?
1.a. ≈ 0.57355 or ≈ 57.355%
b.female; there are more females enrolled.
2.a. ≈ 0.026448 or ≈ 2.64%
b.≈ 0.040302 or ≈ 4.03%
CC-15 55
Using the formula for conditional probability, you can calculate a conditional
probability from other probabilities.
Problem 3
Another approach to Problem 3 is to recognize
that once it is given that the customer is male, the
sample space is restricted to the first row of the
table. To find the size of the new sample space, add
the entries in the first column, 12 + 8 = 20. This is
the total number of males in the sample space. The
number of males who pay online is 12. Dividing,
you get 12
20 = 0.6.
Problem 3 Using the Conditional Probability Formula
Market Research A utility company asked 50 of its customers
whether they pay their bills online or by mail. What is the
probability that a customer pays the bill online, given that the
customer is male?
To use P (online male)
Q Suppose you know that a customer pays by mail. Is
=
it more likely that the customer is male or female?
Explain. ​[14 customers pay by mail. Of these,
Online
By
Mail
Male
12
8
Female
24
6
What are
branches
point?
The tree fir
at “gradua
“not gradu
of these br
“happy” a
happy.”
12
50
20
P (male) =
50
P (male and online) =
P (male and online)
,
P (male)
you need
P(male and online)
and P(male).
eight are male and six are female. Therefore, the
8
probability that a customer is male is 14
? 0.57.
Thus, it is more likely that a customer is male,
given that the customer pays by mail, even
though there are more females in the original
sample space.]
Bill Payment
P (online ∣ male) =
There are 20 males, and
12 males who pay online
out of 50 total customers.
Substitute and simplify.
P (male and online)
P (male)
12
50
= 20
50
=
3
12
= = 0.6
5
20
The probability that a customer pays
online given that the customer is male
is 0.6.
Got It?
Q What two probabilities do you need to find to
Got It? 3. Researchers asked shampoo users
solve this problem? ​[P(female and directly onto
head) and P(female)]
whether they apply shampoo directly to
the head, or indirectly using a hand.
What is the probability that a respondent
applies shampoo directly to the head,
given that the respondent is female?
Q Without using the formula, how would you find
P (Directly on Head 0 Female)? ​[Divide the number
in the female and directly onto head box by the
total in the female row.]
It follows from P (B A) =
P (A and B)
P (A)
Which pa
you follo
Follow the
represents
who are h
their prese
Applying Shampoo
that P (A and B) = P (A)
Directly Onto
Head
Into Hand
First
Male
2
18
Female
6
24
# P (B A).
You can use this rule along with a tree diagram to find probabilities of
dependent events.
56
Chapter 11
Probability and Statistics
56
HSM15_A2Reg_SE_CC_15_TrKit.indd
56
Common Core
HSM15_A2Reg_SE_CC_15_TrKit.ind
02/08/13 3:05 PM
Additional Problems
Answers
HSM15_A2Reg_SE_TrKit.indd Page 56 03/08/13 3:53 PM gg-018
1.The table shows the number
of male and female freshmen
who chose to play one of
the three intramural sports
offered at a small college.
Male
Female
Basketball
54
40
Soccer
36
61
Volleyball
10
12
What is P (soccer 0 female)?
ANSWER a
​ bout 0.54
2.A student compiled the
following table comparing
the amounts of land area and
water area, in square miles, in
various U.S. states.
56 Common Core
Land
Water
Alaska
571,936
91,332
Florida
54,018
11,777
Texas
262,100
6721
California
156,002
7694
What is the probability that a
point chosen at random on a
map is water, given that the
map is of Florida?
ANSWER a
​ bout 17.9%
3.The table shows the
number of male and female
customers at a café on a
certain day. Each customer
drank either a soda or an
iced tea.
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/A ...
Soda
Iced Tea
Male
21
29
Female
28
42
What is the probability that a
customer drank a soda, given
that the customer was male?
ANSWER 0
​ .42
4.The chance of rain for the
next evening is 60%. If it
rains, the chance of lightning
is 80%. If it does not rain,
the chance of lighting is 5%.
What is the probability that it
will not rain and there will be
no lightning?
ANSWER 3
​ 8%
Got It? (continued)
3.0.2
Problem 4 Using a Tree Diagram
Problem 4
Education A school system compiled the following information from a survey it sent
to people who were juniors 10 years earlier.
A tree diagram is an alternative way to represent
a sample space. The advantage of a tree diagram
is that you can mark different probabilities for
each branch; in an ordered-list sample space the
probability of each item in the set is assumed to be
equal and it is the number of combinations of items
that determines probability.
• 85% of the students graduated from high school.
• Of the students who graduated from high school, 90% are happy with their
present jobs.
• Of the students who did not graduate from high school, 60% are happy with
their present jobs.
What are the
branches at each
point?
The tree first branches
at “graduated” and
“not graduated.” Each
of these branches at
“happy” and “not
happy.”
What is the probability that a person from the junior class 10 years ago graduated
from high school and is happy with his or her present job?
Make a tree diagram to help organize the information.
Let G = graduated, NG = not graduated, H = happy with present job,
and NH = not happy with present job.
This branch represents
the students who did
graduate.
G
Each first branch
represents a simple
probability P(G)  0.85
and P(NG)  0.15.
0.85
0.15
NG
0.90
H
0.10
NH
0.60
H
Q The given information in the problem states the
probability a student graduated high school. How
was the probability that a student did not graduate
found? ​[Every student either graduated or did
not graduate, so the total probability must be 1,
1 − 0.85 = 0.15.]
Each second branch
represents a conditional
probability P(H G)  0.90
and P(NH NG)  0.40
Q The tree diagram shows four paths. What do you
think the sum of the probabilities for all four
paths should be? Explain and verify your
answer. ​[The sum of the probabilities for all
0.40
NH
four paths represents everything that could
happen, so the probability should be 100%.
P (H and G) + P (NH and G) + P (H and NG) +
P (NH and NG) = 0.765 + 0.085 + 0.09 +
0.06 = 1.]
This branch represents
the students who did
not graduate.
Which path should
you follow?
Follow the path that
represents graduates
who are happy with
their present job.
The blue highlighted path represents P(G and H).
P (G and H) = P (G)
= 0.85
# P (H G)
# 0.90
Got It?
= 0.765
Q Which of the four paths did you follow to answer
The probability that a person from the junior class 10 years ago graduated and is happy
with his or her present job is 0.765, or 76.5%.
this question? What is the probability of each
section of the path? ​[The path is “not graduated”
Got It? 4. What is the probability that a student from the junior class 10 years ago in
and “is happy.” P (NG) = 0.15 and P (H ∣ NG) = 0.6]
Problem 4 did not graduate and is happy with his or her present job?
Q If the graduating class contained 600 students,
how many students would you expect did not
graduate and are happy with their present jobs? ​
[54 students]
Lesson 11-4
HSM15_A2Reg_SE_CC_15_TrKit.indd
57
02/08/13 3:05 PM
Conditional Probability
CC-15 Conditional Probability
57
57
02/08/13 3:05 PM
4.9%
HSM15_A2Reg_SE_TrKit.indd Page 57 03/08/13 3:53 PM gg-018
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/A ...
CC-15 57
3 ​Lesson Check
Do you know HOW?
•For Exercises 1–3, students must understand that
it is given that the drawn card is black. That fact
does not need to be calculated. Therefore, the
denominator for calculating these probabilities
should be 26 rather than 52.
•For Exercise 4, students may benefit from
drawing a tree diagram. They should determine
that the first branch is for illustrated/notillustrated and the second branches are for
hardback/not-hardback.
Lesson Check
Do you know HOW?
2. P (4)
3. P(diamond)
8. P(has diploma)
•In Exercise 5, students may think that the sum
of the probabilities of paired branches should
equal the probability of the preceding branch in
the diagram. Explain that the smaller branches
contain 100% (all) of the preceding branch,
which is the same as a part of the entire diagram.
•For Exercise 6, the situation must have at least
two choices or events, and the first should be
given.
9. P(has diploma and experience)
App
MATHEMATICAL
PRACTICES
See Problems 1 and 2.
Use the table to find each probability.
Do you UNDERSTAND?
B
7. Compare and Contrast How are the Fundamental
Counting Principle and tree diagrams alike? How are
they different?
Practice and Problem-Solving Exercises
Practice
6. Open-Ended Describe a situation in which you
would use conditional probability to find the answer.
4. The probability that a car has two doors, given that it
is red, is 0.6. The probability that a car has two doors
and is red is 0.2. What is the probability that a car is
red?
A
PRACTICES
5. Reasoning Using the tree diagram in Problem 4,
explain why the probabilities on each pair of
branches must add up to 1.
A card is drawn from a standard deck of cards. Find
each probability, given that the card drawn is black.
1. P(club)
MATHEMATICAL
Do you UNDERSTAND?
Characteristics of Job Applicants
Has Experience
10. P(has experience has diploma)
Has High
School Diploma
11. P(has no diploma has experience)
Yes
No
54
5
27
4
Yes
No
Use the table to find each probability.
12. P(The recipient is male.)
Projected Number of Degree
Recipients in 2010 (thousands)
13. P(The degree is a bachelor’s.)
14. P(The recipient is female, given that the
degree is an associate’s.)
Close
15. P(The degree is not an associate’s, given
that the recipient is male.)
Q What is a simple probability experiment that shows
how the probability of an outcome changes when
a part is given? ​[Answers may vary. Sample:
Degree
Male
Female
Associate’s
245
433
Bachelor’s
598
858
SOURCE: U.S. National Center for Education Statistics
See Problems 3 and 4.
Use the survey results for Exercises 16 and 17.
16. Find the probability that a respondent has a pet,
given that the respondent has had a pet.
Flipping a coin twice and getting two heads has
a probability of one fourth. If the first coin flip is
given as heads, then the probability of getting
two heads is one half. The probability doubled.]
C
Cha
39% have a pet now and have had a pet.
61% do not have a pet now.
17. Find the probability that a respondent has never
had a pet, given that the respondent does not have
a pet now.
86% have had a pet.
14% do not have a pet now and have
never had a pet.
18. Sports A football team has a 70% chance
of winning when it doesn’t snow, but only a
40% chance of winning when it snows. Suppose there is a 50% chance of snow.
Make a tree diagram to find the probability that the team will win.
58
Chapter 11
Probability and Statistics
58
HSM15_A2Reg_SE_CC_15_TrKit.indd
Common58Core
Answers
HSM15_A2Reg_SE_TrKit.indd Page 58 8/13/13 12:29 AM epg
Lesson Check
1.1
2
1
2.13
, or about 7.7%
3.0%
4.50%
5.The sum of the probability of an event
happening and the probability of an
event not happening is 1. Each branch
represents either the event happening
or the event not happening.
6.Check students’ work.
7.Answers may vary. Sample: Tree
diagrams apply to cases in which
more than one event occurs in a
sequence. The Fundamental Counting
Principle applies to situations in which
there are multiple outcomes of a
single event. With a tree diagram,
58 Common Core
HSM15_A2Reg_SE_CC_15_TrKit.ind
02/08/13 3:05 PM
but not with the Fundamental
Counting Principle, you can determine
probabilities of dependent events, or
conditional probabilities.
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/S ...
Practice and Problem-Solving
Exercises
8.0.99.0.6
10.0.611. ≈ 0.085
12. ≈ 0.39513. ≈ 0.682
14. ≈ 0.63915. ≈ 0.709
16. ≈ 45%17. ≈ 23%
18. 0. 4 W
0. 5
0. 5
snow
no s
now
0. 6 L
0.7 W
0. 3 L
P(W) = 55%
W  win
L  loss
19. Make a tree diagram based on the survey results below. Then find P(a female
respondent is left-handed) and P(a respondent is both male and right-handed).
• Of all the respondents, 17% are male.
• Of the male respondents, 33% are left-handed.
• Of female respondents, 90% are right-handed.
B
Apply
e
nd 4.
ASSIGNMENT GUIDE
Basic: 8–19 all, 20–26
20. Suppose A and B are independent events, with P (A) = 0.60 and P (B) = 0.25. Find
each probability.
a. P (A and B)
b. P (A B)
c. What do you notice about P (A) and P(A B)?
d. Reasoning One way to describe A and B as independent events is The
occurrence of B has no effect on the probability of A. Explain how the answer to
part (c) illustrates this relationship.
r.
nd 2.
4 ​Practice
Average: 9–19 odd, 20–31
Advanced: 9–19 odd, 20–33
Mathematical Practices are supported by
exercises with red headings. Here are the Practices
supported in this lesson:
21. Think About a Plan A math teacher gives her class two tests. 60% of the class
passes both tests and 80% of the class passes the first test. What percent of those
who pass the first test also pass the second test?
• What conditional probability are you looking for?
• How can a tree diagram help you solve this problem?
MP 1: Make Sense of Problems Ex. 21
MP 2: Reason Quantitatively Ex. 32a
MP 3: Construct Arguments Ex. 5, 6
MP 3: Communicate Ex. 5, 20d, 32a
MP 3: Compare Arguments Ex. 7
MP 4: Model with Mathematics Ex. 33
Weather Use probability notation to describe the chance of each event. Let
S, C, W, and R represent sunny, cloudy, windy, and rainy weather, respectively.
22. cloudy weather
23. sunny and windy weather
24. rainy weather if it is windy
Applications exercises have blue headings.
Exercises 18 and 25 support MP 4: Model.
25. Transportation You can take Bus 65 or Bus 79. You take the first bus that arrives.
The probability that Bus 65 arrives first is 75%. There is a 40% chance that Bus 65
picks up passengers along the way. There is a 60% chance that Bus 79 picks up
passengers. Your bus picked up passengers. What is the probability that it was Bus 65?
The tree diagram relates snowfall and school closings. Find each probability.
Let H, L, O, and C represent heavy snowfall, light snowfall, schools open, and
schools closed, respectively.
C
Challenge
et.
26. P(C)
27. P(H and O)
28. P (H C)
29. P (L O)
30. P (L C)
31. P (H O)
0.8
H
0.4
To check students’ understanding of key skills and
concepts, go over Exercises 9, 17, 20, 21, and 25.
0.2
0.3
0.6
L
0.7
32. a. Writing Explain which branches of the tree diagram at the right represent
A
conditional probabilities. Give a specific example.
b. Are the event of having a license and the event of being an adult
independent events? Justify your answer.
M
c. Open-Ended Estimate probabilities for each branch of the tree diagram
A  adult (21 or older)
for your city or town. Then find P(L).
M  minor (under 21)
33. Reasoning Sixty percent of a company’s sales representatives have
L  licensed driver
completed training seminars. Of these, 80% have had increased sales.
N  not licensed to drive
Overall, 56% of the representatives (whether trained or not) have had
increased sales. Use a tree diagram to find the probability of increased sales,
given that a representative has not been trained.
Lesson 11-4
33 L
M 0.
0.67 R
0.1 L
0. 8 3
F 0.9
R
0.17
M
F
R
L
HSM15_A2Reg_SE_TrKit.indd Page 59 03/08/13 3:53 PM gg-018
 male
 female
 right-handed
 left-handed
P (L 0 F ) = 10%, P(M and R) ≈ 11.4%
20. a. 0.15
b.0.60
c–d. Since P (A) = P (A 0 B), the
probability of A is the same,
regardless of the occurrence of B.
21. 75%
22. P (C )
23. P (S and W )
24. P (R 0 W )
25. 2
, or 66.67%
3
26. 0.50, or 50%
27. 0.08, or 8%
28. 0.64
29. 0.84
30. 0.36
31. 0.16
C
O
C
O
L
N
L
N
59
Conditional Probability
59
CC-15 Conditional Probability
HSM15_A2Reg_SE_CC_15_TrKit.indd
59
02/08/13 3:05 PM
19.
HOMEWORK QUICK CHECK
02/08/13 3:05 PM
32. a. The branches labelled L and N
represent conditional probabilities
dependent upon the person being
an adult or a minor. For example, the
top branch represents the probability
that a person is licensed given that
he or she is an adult.
b.No; the probability of a minor
being licensed is not the same as
the probability of an adult being
licensed.
c.Check students’ work.
8 I
33.
T 0.
/120/PE01457/TRANSITION_KITS/NA/ANCILLARY/2015/XXXXXXXXXX/Layout/Interior_Files/A ...
0. 6
0. 4
0. 2
N
0. 2 I
R 0. 8
N
T representative that completed training seminars
R representative that didn’t complete a training seminar
I representative with increased sales
N representative without increased sales
P (I 0 N) = 0.2
CC-15 59