Transcript File

+
Chapter 5: Probability: What are the Chances?
Section 5.3
Conditional Probability and Independence
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
+
Chapter 5
Probability: What Are the Chances?
 5.1
Randomness, Probability, and Simulation
 5.2
Probability Rules
 5.3
Conditional Probability and Independence
+Section 5.3
Conditional Probability and Independence
Learning Objectives
After this section, you should be able to…

DEFINE conditional probability

COMPUTE conditional probabilities

DESCRIBE chance behavior with a tree diagram

DEFINE independent events

DETERMINE whether two events are independent

APPLY the general multiplication rule to solve probability questions
is Conditional Probability?
When we are trying to find the probability that one event will happen
under the condition that some other event is already known to have
occurred, we are trying to determine a conditional probability.
Definition:
The probability that one event happens given that another event
is already known to have happened is called a conditional
probability. Suppose we know that event A has happened.
Then the probability that event B happens given that event A
has happened is denoted by P(B | A).
Read | as “given that”
or “under the
condition that”
Conditional Probability and Independence
The probability we assign to an event can change if we know that some
other event has occurred. This idea is the key to many applications of
probability.
+
 What
Example: Who Owns a Home
Yes
No
Total
Homeowner
221
119
340
Not a Homeowner
89
71
160
Total
310
190
500
1. If we know that a person owns a home, what is the probability that the
person is a high school graduate? There are a total of 340 people in the
sample that own a home. Because there are 221 high school graduates
among the 340 home owners, the desired probability is
P(is a high school graduate given owns a home) = 221/340 or 65%
2. If we know that a person is a high school graduate, what is the probability
that the person owns a home? There are a total of 310 people who are
high school graduates. Because there are 221 home owners among the
310 high school graduates, the desired probability is
P(owns a home given is a high school graduate) = 221/310 or about 71%
Conditional Probability and Independence
High School Graduate?
+
 Alternate
Grade Distributions
+
 Example:
E: the grade comes from an EPS course, and
L: the grade is lower than a B.
Total
6300
1600
2100
Total 3392 2952
Find P(L)
P(L) = 3656 / 10000 = 0.3656
Find P(E | L)
P(E | L) = 800 / 3656 = 0.2188
Find P(L | E)
P(L| E) = 800 / 1600 = 0.5000
3656
10000
Conditional Probability and Independence
Consider the two-way table on page 314. Define events
Probability and Independence
Definition:
Two events A and B are independent if the occurrence of one
event has no effect on the chance that the other event will
happen. In other words, events A and B are independent if
P(A | B) = P(A) and P(B | A) = P(B).
Example:
Are the events “male” and “left-handed”
independent? Justify your answer.
P(left-handed | male) = 3/23 = 0.13
P(left-handed) = 7/50 = 0.14
These probabilities are not equal, therefore the
events “male” and “left-handed” are not independent.
Conditional Probability and Independence
When knowledge that one event has happened does not change
the likelihood that another event will happen, we say the two
events are independent.
+
 Conditional
Is there a relationship between gender and having allergies?
To find out, we used the random sampler at the United States
Census at School website (www.amstat.org/censusatschool)
to randomly select 40 US high school students who completed
a survey. The two-way table shows the gender of each
student and whether the student has allergies.
Allergies
No Allergies
Total
Female
10
13
23
Male
8
9
17
Total
18
22
40
P(allergies | female) = 10/23 = 0.435
Are the events “female” and
“allergies” independent? Justify your
answer.
P(allergies) = 18/40 = 0.45
These probabilities are close but not equal, therefore the events
“female” and “allergies” are not independent. Knowing that a student
was female slightly lowered the probability that she has allergies.
Conditional Probability and Independence

Example: Allergies
+
 Alternate
Diagrams
Consider flipping a
coin twice.
What is the probability
of getting two heads?
Sample Space:
HH HT TH TT
So, P(two heads) = P(HH) = 1/4
Conditional Probability and Independence
We learned how to describe the sample space S of a chance
process in Section 5.2. Another way to model chance
behavior that involves a sequence of outcomes is to construct
a tree diagram.
+
 Tree
Two Sneezers
Allergies
No Allergies
Total
Male
8
9
17
Total
18
22
40
Suppose we chose 2 students
at random. What is the
probability that both students
suffer from allergies?
To get two students who suffer from allergies, we need to get an allergy sufferer
for the first student and an allergy sufferer for the second student. Following
along the top branches of the tree, we see that the probability is:
P(two allergy sufferers) = P(1st student has allergies and 2nd student has
allergies)
= P(1st student has allergies) P(2nd student has allergies | 1st student has
allergies)
= (18/40)(17/39) = 0.196
There is about a 20% chance of selecting two students with allergies.
Conditional Probability and Independence
In the previous alternate
example, we used a two-way
table that classified 40 students
according to their gender and
whether they had allergies.
Here is the table again.
Female
10
13
23
+
 Picking
Multiplication Rule
General Multiplication Rule
The probability that events A and B both occur can be
found using the general multiplication rule
P(A ∩ B) = P(A) • P(B | A)
where P(B | A) is the conditional probability that event
B occurs given that event A has already occurred.
Conditional Probability and Independence
The idea of multiplying along the branches in a tree diagram
leads to a general method for finding the probability P(A ∩ B)
that two events happen together.
+
 General
Teens with Online Profiles
What percent of teens are online and have posted a profile?
P(online )  0.93
P(profile | online )  0.55
P(online and have profile )  P(online ) P(profile | online )


 (0.93)(0.55)
 0.5115
51.15% of teens are online and have
 a profile.
posted
Conditional Probability and Independence
The Pew Internet and American Life Project finds that 93% of teenagers (ages
12 to 17) use the Internet, and that 55% of online teens have posted a profile
on a social-networking site.
+
 Example:
About 55% of high school students participate in a school athletic team at
some level and about 5% of these athletes go on to play on a college team in
the NCAA.
What percent of high school students play a sport in high school and go
on to play a sport in the NCAA?
P(high school sport )  0.55
P( NCAA sport | high school sport )  0.05
P(high school sport and NCAA sport )
 P(high school sport )  P( NCAA sport | high school sport )
 (0.55)(0.05)  0.0275
Almost 3% of high school students will play a sport in high school
and in the NCAA.
Conditional Probability and Independence

Playing in the NCAA
+
 Example:
Who Visits YouTube?
See the example on page 320 regarding adult Internet users.
What percent of all adult Internet users visit video-sharing sites?
P(video yes ∩ 18 to 29) = 0.27 • 0.7
=0.1890
P(video yes ∩ 30 to 49) = 0.45 • 0.51
=0.2295
P(video yes ∩ 50 +) = 0.28 • 0.26
=0.0728
P(video yes) = 0.1890 + 0.2295 + 0.0728 = 0.4913
+
 Example:
Media Usage and Good Grades
+
 Example:
In a study, 17% of the youth were classified as light media users, 62% were classified as
moderate media users and 21% were classified as heavy media users. Of the light users
who responded, 74% described their grades as good (A’s and B’s), while only 68% of the
moderate users and 52% of the heavy users described their grades as good. According to
this study, what percent of young people ages 8-18 described their grades as good?
State: What percent of young people get good grades?
Plan: If we choose a subject in the
study at random, P(light user) =
0.17, P(moderate user) = 0.62,
P(heavy user) = 0.21, P(good
grades | light user) = 0.74, P(good
grades | moderate user) = 0.68,
P(good grades | heavy user) = 0.52.
We want to find the unconditional
probability P(good grades). A tree
diagram should help.
Do: There are three groups of students who get good grades, those who are light users and
get good grades, those who are moderate users and get good grades, and those who are
heavy users and get good grades. Because these groups are mutually exclusive, we can
add the probabilities of being in one of these three groups. P(good grades) = (0.17)(0.74) +
(0.62)(0.68) + (0.21)(0.52) = 0.1258 + 0.4216 + 0.1092 = 0.6566.
Conclude: About 66% of the students in the study described their grades as good.
A Special Multiplication Rule
Definition:
Multiplication rule for independent events
If A and B are independent events, then the probability that A
and B both occur is
P(A ∩ B) = P(A) • P(B)
Example:
Following the Space Shuttle Challenger disaster, it was determined that the failure
of O-ring joints in the shuttle’s booster rockets was to blame. Under cold
conditions, it was estimated that the probability that an individual O-ring joint would
function properly was 0.977. Assuming O-ring joints succeed or fail independently,
what is the probability all six would function properly?
P(joint1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK)
=P(joint 1 OK) • P(joint 2 OK) • … • P(joint 6 OK)
=(0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87
Conditional Probability and Independence
When events A and B are independent, we can simplify the
general multiplication rule since P(B| A) = P(B).
+
 Independence:
First Trimester Screen
+
 Example:
The First Trimester Screen is a non-invasive test given during the first trimester of
pregnancy to determine if there are specific chromosomal abnormalities in the fetus.
According to a study published in the New England Journal of Medicine in November 2005
(http://www.americanpregnancy.org/prenataltesting/firstscreen.html), approximately 5% of
normal pregnancies will receive a positive result. Among 100 women with normal
pregnancies, what is the probability that there will be at least one false positive?
State: If 100 women with normal pregnancies are tested with the First Trimester
Screen, what is the probability that at least one woman will receive a positive
result?
Plan: It is reasonable to assume that the test results for different women are
independent. To find the probability of at least one false positive, we can use
the complement rule and the probability that none of the women will receive a
positive test result.
P(at least one positive) = 1 – P(no positive results)
Do: For women with normal pregnancies, the probability that a single test is not
positive is 1 – 0.05 = 0.95. The probability that all 100 women will get negative
results is (0.95)(0.95) (0.95) = (0.95)100 = 0.0059. Thus, P(at least one positive) =
1 – 0.0059 = 0.9941.
Conclude: There is over a 99% probability that at least one of the 100 women with
normal pregnancies will receive a false positive on the First Trimester Screen.
Conditional Probabilities
General Multiplication Rule
P(A ∩ B) = P(A) • P(B | A)
Conditional Probability Formula
To find the conditional probability P(B | A), use the formula
=
Conditional Probability and Independence
If we rearrange the terms in the general multiplication rule, we
can get a formula for the conditional probability P(B | A).
+
 Calculating
Who Reads the Newspaper?
What is the probability that a randomly selected resident who reads USA
Today also reads the New York Times?
P(A  B)
P(B | A) 
P(A)
P(A  B)  0.05


P(A)  0.40
0.05
P(B | A) 
 0.125
0.40
There is a 12.5% chance that a randomly selected resident who reads USA
Today also reads the New York Times.
Conditional Probability and Independence
In Section 5.2, we noted that residents of a large apartment complex can be
classified based on the events A: reads USA Today and B: reads the New
York Times. The Venn Diagram below describes the residents.
+
 Example:
In an alternate example in section 5.2, we classified US households according
to the types of phones they used.
What is the probability that a randomly selected household with a landline
also has a cell phone?
Landline
No
Landline
Total
Cell
Phone
0.60
No Cell
Phone
0.18
0.20
0.02
0.22
0.80
0.20
1.00
Total
0.78

We want to find P(cell phone | landline).
Using the conditional probability
formula,
P(cellphone and landline )
P(cellphone | landline ) 
P(landline )
0.60

 0.77
0.78
There is a 77% chance that a landline user also has a cell
phone.
Conditional Probability and Independence

Who Reads the Newspaper?
+
 Example:
+ Section 5.3
Conditional Probability and Independence
Summary
In this section, we learned that…

If one event has happened, the chance that another event will happen is a
conditional probability. P(B|A) represents the probability that event B
occurs given that event A has occurred.

Events A and B are independent if the chance that event B occurs is not
affected by whether event A occurs. If two events are mutually exclusive
(disjoint), they cannot be independent.

When chance behavior involves a sequence of outcomes, a tree diagram
can be used to describe the sample space.

The general multiplication rule states that the probability of events A
and B occurring together is P(A ∩ B)=P(A) • P(B|A)

In the special case of independent events, P(A ∩ B)=P(A) • P(B)

The conditional probability formula states P(B|A) = P(A ∩ B) / P(A)
+
Looking Ahead…
In the next Chapter…
We’ll learn how to describe chance processes using the
concept of a random variable.
We’ll learn about
 Discrete and Continuous Random Variables
 Transforming and Combining Random Variables
 Binomial and Geometric Random Variables