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Chapter
5
Probability
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Section
5.1
Probability Rules
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Apply the rules of probabilities
2. Compute and interpret probabilities using the
empirical method
3. Compute and interpret probabilities using the
classical method
4. Use simulation to obtain data based on
probabilities
5. Recognize and interpret subjective
probabilities
5-3
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Probability is a measure of the likelihood of a
random phenomenon or chance behavior.
Probability describes the long-term proportion
with which a certain outcome will occur in
situations with short-term uncertainty.
Use the probability applet to simulate flipping a
coin 100 times. Plot the proportion of heads
against the number of flips. Repeat the
simulation.
5-4
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Probability deals with experiments that yield
random short-term results or outcomes, yet
reveal long-term predictability.
The long-term proportion in which a
certain outcome is observed is the
probability of that outcome.
5-5
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The Law of Large Numbers
As the number of repetitions of a probability
experiment increases, the proportion with
which a certain outcome is observed gets
closer to the probability of the outcome.
5-6
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In probability, an experiment is any process that
can be repeated in which the results are
uncertain.
The sample space, S, of a probability experiment
is the collection of all possible outcomes.
An event is any collection of outcomes from a
probability experiment. An event may consist of
one outcome or more than one outcome.
We will denote events with one outcome,
sometimes called simple events, ei. In general,
events are denoted using capital letters such as E.
5-7
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Experiment
Single Birth
3 births
Example of event
1 boy (single event)
2 G and 1 B
not simple event
Sample Space
{g,b}
{bbb,bbg,bgb,gbb,
ggg,ggb,gbg,bgg}
2 G and 1 B is not a simple event because it can be broken down
into simpler events like ggb, gbg or bgg (simple events)
5-8
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EXAMPLE
Identifying Events and the Sample Space of a
Probability Experiment
Consider the probability experiment of having two
children.
(a) Identify the outcomes of the probability experiment.
(b) Determine the sample space.
(c) Define the event E = “have one boy”.
(a) e1 = boy, boy, e2 = boy, girl, e3 = girl, boy, e4 = girl, girl
(b) {(boy, boy), (boy, girl), (girl, boy), (girl, girl)}
(c) {(boy, girl), (girl, boy)}
5-9
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Objective 1
• Apply Rules of Probabilities
5-10
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Rules of probabilities
1.
The probability of any event E,
P(E),
must be greater than or equal to
0
and less than or equal to 1.
That is, 0 ≤ P(E) ≤ 1.
2.
all
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The sum of the probabilities of
outcomes must equal 1.
That is, if the sample space
S = {e1, e2, …, en}, then
P(e1) + P(e2) + … + P(en) = 1
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A probability model lists the possible
outcomes of a probability experiment and
each outcome’s probability. A
probability model must satisfy rules 1
and 2 of the rules of probabilities.
5-12
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EXAMPLE
A Probability Model
In a bag of peanut M&M milk
chocolate candies, the colors of
the candies can be brown, yellow,
red, blue, orange, or green.
Suppose that a candy is randomly
selected from a bag. The table
shows each color and the
probability of drawing that color.
Verify this is a probability model.
Color
Probability
Brown
0.12
Yellow
0.15
Red
0.12
Blue
0.23
Orange
0.23
Green
0.15
• All probabilities are between 0 and 1, inclusive.
• Because 0.12 + 0.15 + 0.12 + 0.23 + 0.23 + 0.15 = 1, rule
2 (the sum of all probabilities must equal 1) is satisfied.
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If an event is impossible, the probability
of the event is 0.
If an event is a certainty, the
probability of the event is 1.
An unusual event is an event that has
a low probability of occurring.
5-14
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Objective 2
• Compute and Interpret Probabilities Using the
Empirical Method
5-15
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Approximating Probabilities Using the
Empirical Approach
The probability of an event E is
approximately the number of times event
E is observed divided by the number of
repetitions of the experiment.
P(E) ≈ relative frequency of E
frequency of E

number of trials of experiment
5-16
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EXAMPLE
Building a Probability Model
Pass the PigsTM is a MiltonBradley game in which pigs
are used as dice. Points are
earned based on the way
the pig lands. There are six
possible outcomes when
one pig is tossed. A class of
52 students rolled pigs
3,939 times. The number of
times each outcome
occurred is recorded in the
table at right
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Outcome
Side with no dot
Side with dot
Frequency
1344
1294
Razorback
Trotter
Snouter
767
365
137
Leaning Jowler
32
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EXAMPLE
Building a Probability Model
Outcome
Frequency
(a) Use the results of the
Side with no dot
1344
experiment to build a
probability model for
Side with dot
1294
the way the pig lands. Razorback
767
(b) Estimate the
Trotter
365
probability that a
Snouter
137
thrown pig lands on
Leaning Jowler
32
the “side with dot”.
(c) Would it be unusual to throw a “Leaning Jowler”?
5-18
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(a)
Outcome
Side with no dot
5-19
Probability
1344
 0.341
3939
Side with dot
0.329
Razorback
0.195
Trotter
0.093
Snouter
0.035
Leaning Jowler
0.008
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(a)
Outcome
Side with no dot
Side with dot
Razorback
Trotter
Snouter
Leaning Jowler
Probability
0.341
0.329
0.195
0.093
0.035
0.008
(b) The probability a throw results in a “side with dot”
is 0.329. In 1000 throws of the pig, we would
expect about 329 to land on a “side with dot”.
5-20
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(a)
Outcome
Side with no dot
Side with dot
Razorback
Trotter
Snouter
Leaning Jowler
Probability
0.341
0.329
0.195
0.093
0.035
0.008
(c) A “Leaning Jowler” would be unusual. We would
expect in 1000 throws of the pig to obtain
“Leaning Jowler” about 8 times.
5-21
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Objective 3
• Compute and Interpret Probabilities Using the
Classical Method
5-22
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The classical method of computing
probabilities requires equally likely
outcomes.
An experiment is said to have equally
likely outcomes when each simple event
has the same probability of occurring.
5-23
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Computing Probability Using the Classical
Method
If an experiment has n equally likely
outcomes and if the number of ways that an
event E can occur is m, then the probability
of E, P(E) is
number of ways E can occur m
P E  

number of possible outcomes n
5-24
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Computing Probability Using the Classical
Method
So, if S is the sample space of this
experiment,
N E 
P E  
N S 
where N(E) is the number of outcomes in E, and
N(S) is the number of outcomes in the sample
space.
5-25
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EXAMPLE
Computing Probabilities Using the Classical Method
Suppose a “fun size” bag of M&Ms contains 9 brown
candies, 6 yellow candies, 7 red candies, 4 orange
candies, 2 blue candies, and 2 green candies. Suppose
that a candy is randomly selected.
(a) What is the probability that it is yellow?
(b) What is the probability that it is blue?
(c) Comment on the likelihood of the candy being
yellow versus blue.
5-26
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EXAMPLE
Computing Probabilities Using the Classical Method
(a)There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30
candies, so N(S) = 30.
N(yellow)
P(yellow) 
N (S)
6

 0.2
30
(b) P(blue) = 2/30 = 0.067.
(c) Since P(yellow) = 6/30 and P(blue) = 2/30,
selecting a yellow is three times as likely as
selecting a blue.
5-27
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Objective 4
• Use Simulation to Obtain Data Based on
Probabilities
5-28
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EXAMPLE
Using Simulation
Use the probability applet to simulate throwing a 6sided die 100 times. Approximate the probability of
rolling a 4. How does this compare to the classical
probability? Repeat the exercise for 1000 throws of the
die.
5-29
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Objective 5
• Recognize and Interpret Subjective
Probabilities
5-30
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The subjective probability of an outcome
is a probability obtained on the basis of
personal judgment.
For example, an economist predicting
there is a 20% chance of recession next
year would be a subjective probability.
5-31
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EXAMPLE
Empirical, Classical, or Subjective Probability
In his fall 1998 article in Chance Magazine, (“A
Statistician Reads the Sports Pages,” pp. 17-21,) Hal
Stern investigated the probabilities that a particular
horse will win a race. He reports that these
probabilities are based on the amount of money bet
on each horse. When a probability is given that a
particular horse will win a race, is this empirical,
classical, or subjective probability?
Subjective because it is based upon people’s
feelings about which horse will win the race.
The probability is not based on a probability
experiment or counting equally likely outcomes.
5-32
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Section
5.2
The Addition Rule
and
Complements
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Objectives
1. Use the Addition Rule for Disjoint Events
2. Use the General Addition Rule
3. Compute the probability of an event using
the Complement Rule
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Objective 1
• Use the Addition Rule for Disjoint Events
5-35
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Two events are disjoint if they have no
outcomes in common. Another name for
disjoint events is mutually exclusive
events.
5-36
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We often draw pictures of events using Venn diagrams.
These pictures represent events as circles enclosed in a
rectangle. The rectangle represents the sample space, and
each circle represents an event. For example, suppose we
randomly select a chip from a bag where each chip in the
bag is labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Let E represent
the event “choose a number less than or equal to 2,” and
let F represent the event “choose a number greater than or
equal to 8.” These events are disjoint as shown in the
figure.
5-37
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Addition Rule for Disjoint Events
If E and F are disjoint (or mutually
exclusive) events, then
P E or F   P E  P F 
5-39
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The Addition Rule for Disjoint Events
can be extended to more than two
disjoint events.
In general, if E, F, G, . . . each have no
outcomes in common (they are
pairwise disjoint), then
P E or F or G or ...  P E  P F  P G L
5-40
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EXAMPLE
The Addition Rule for Disjoint Events
The probability model to
the right shows the
distribution of the number
of rooms in housing units
in the United States.
(a) Verify that this is a
probability model.
All probabilities are
between 0 and 1, inclusive.
Number of Rooms Probability
in Housing Unit
One
0.010
Two
0.032
Three
0.093
Four
Five
Six
Seven
Eight
Nine or more
0.010 + 0.032 + … + 0.080 = 1
5-41
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0.176
0.219
0.189
0.122
0.079
0.080
EXAMPLE
The Addition Rule for Disjoint Events
(b) What is the
probability a randomly
selected housing unit
has two or three
rooms?
P(two or three)
= P(two) + P(three)
= 0.032 + 0.093
= 0.125
5-42
Number of Rooms Probability
in Housing Unit
One
0.010
Two
0.032
Three
0.093
Four
Five
Six
Seven
Eight
Nine or more
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0.176
0.219
0.189
0.122
0.079
0.080
EXAMPLE
The Addition Rule for Disjoint Events
(c) What is the probability
a randomly selected
housing unit has one or
two or three rooms?
Number of Rooms Probability
in Housing Unit
One
0.010
Two
0.032
Three
0.093
P(one or two or three)
Four
Five
= P(one) + P(two) + P(three)
Six
= 0.010 + 0.032 + 0.093
Seven
Eight
= 0.135
Nine or more
5-43
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0.176
0.219
0.189
0.122
0.079
0.080
Objective 2
• Use the General Addition Rule
5-44
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The General Addition Rule
For any two events E and F,
P E or F   P E  P F  P E and F 
5-45
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EXAMPLE
Illustrating the General Addition Rule
Suppose that a pair of dice are thrown. Let E = “the first
die is a two” and let F = “the sum of the dice is less than
or equal to 5”. Find P(E or F) using the General Addition
Rule.
5-46
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N(E)
P(E) 
N(S)
6

36
1

6
N(F)
P(F) 
N(S)
10

36
5

18
P(E and F
N(E and F)

N(S)
3

36
1

12
P(E or F)  P(E)  P(F)  P(E and F)
6 10 3



36 36 36
13

36
5-47
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Objective 3
• Compute the Probability of an Event Using the
Complement Rule
5-48
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Complement of an Event
Let S denote the sample space of a
probability experiment and let E denote
an event. The complement of E, denoted
EC, is all outcomes in the sample space S
that are not outcomes in the event E.
5-49
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Complement Rule
If E represents any event and EC represents
the complement of E, then
P(EC) = 1 – P(E)
Entire region
The area outside
the circle
represents Ec
5-50
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EXAMPLE
Illustrating the Complement Rule
According to the American Veterinary Medical
Association, 31.6% of American households own
a dog. What is the probability that a randomly
selected household does not own a dog?
P(do not own a dog) = 1 – P(own a dog)
= 1 – 0.316
= 0.684
5-51
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EXAMPLE
Computing Probabilities Using Complements
The data to the right
represent the travel time to
work for residents of
Hartford County, CT.
(a) What is the probability a
randomly selected resident
has a travel time of 90 or
more minutes?
Source: United States Census Bureau
5-52
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EXAMPLE
Computing Probabilities Using Complements
There are a total of
24,358 + 39,112 + … + 4,895
= 393,186
residents in Hartford County
The probability a randomly
selected resident will have
a commute time of “90 or
more minutes” is
4895
 0.012
393,186
5-53
Source: United States Census Bureau
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
(b) Compute the probability that a randomly
selected resident of Hartford County, CT will
have a commute time less than 90 minutes.
P(less than 90 minutes) = 1 – P(90 minutes or more)
= 1 – 0.012
= 0.988
5-54
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Section
5.3
Independence
and the
Multiplication
Rule
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Objectives
1. Identify independent events
2. Use the Multiplication Rule for Independent
Events
3. Compute at-least probabilities
5-56
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Objective 1
1. Identify Independent Events
5-57
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Two events E and F are independent if
the occurrence of event E in a
probability experiment does not affect
the probability of event F. Two events
are dependent if the occurrence of event
E in a probability experiment affects the
probability of event F.
5-58
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EXAMPLE Independent or Not?
(a) Suppose you draw a card from a standard 52-card deck
of cards and then roll a die. The events “draw a heart”
and “roll an even number” are independent because the
results of choosing a card do not impact the results of
the die toss.
(b) Suppose two 40-year old women who live in the
United States are randomly selected. The events
“woman 1 survives the year” and “woman 2 survives
the year” are independent.
5-59
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Objective 2
• Use the Multiplication Rule for Independent
Events
5-60
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Multiplication Rule for Independent Events
If E and F are independent events, then
P E and F   P E  P F 
5-61
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EXAMPLE
Computing Probabilities of Independent Events
The probability that a randomly selected female aged 60
years old will survive the year is 99.186% according to
the National Vital Statistics Report, Vol. 47, No. 28.
What is the probability that two randomly selected 60
year old females will survive the year?
The survival of the first female is independent of the
survival of the second female. We also have that
P(survive) = 0.99186.
P First survives and second survives 
 P First survives  P Second survives 
 (0.99186)(0.99186)
 0.9838
5-62
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EXAMPLE
Computing Probabilities of Independent Events
A manufacturer of exercise equipment knows that
10% of their products are defective. They also
know that only 30% of their customers will actually
use the equipment in the first year after it is
purchased. If there is a one-year warranty on the
equipment, what proportion of the customers will
actually make a valid warranty claim?
We assume that the defectiveness of the equipment
is independent of the use of the equipment. So,
P defective and used   P defective  P used 
 (0.10)(0.30)
 0.03
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Multiplication Rule for n Independent Events
If E1, E2, E3, … and En are independent
events, then
P E1 and E2 and E3 and ... and En 
 P E1  P E2    P En 
5-64
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EXAMPLE
Illustrating the Multiplication Principle for Independent Events
The probability that a randomly selected female aged 60
years old will survive the year is 99.186% according to
the National Vital Statistics Report, Vol. 47, No. 28.
What is the probability that four randomly selected 60
year old females will survive the year?
5-65
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EXAMPLE
Illustrating the Multiplication Principle for Independent Events
P(all 4 survive)
= P(1st survives and 2nd survives and 3rd survives and 4th
survives)
= P(1st survives) . P(2nd survives) . P(3rd survives) . P(4th
survives)
= (0.99186) (0.99186) (0.99186) (0.99186)
= 0.9678
5-66
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Objective 3
• Compute At-Least Probabilities
5-67
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EXAMPLE
Computing “at least” Probabilities
The probability that a randomly selected female aged
60 years old will survive the year is 99.186%
according to the National Vital Statistics Report, Vol.
47, No. 28. What is the probability that at least one of
500 randomly selected 60 year old females will die
during the course of the year?
P(at least one dies) = 1 – P(none die)
= 1 – P(all survive)
= 1 – (0.99186)500
= 0.9832
5-68
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Summary: Rules of Probability
1.The probability of any event must be between 0
and 1. If we let E denote any event,
then 0 ≤ P(E) ≤ 1.
2.The sum of the probabilities of all outcomes in
the sample space must equal 1.
That is, if the sample space S = {e1, e2, …, en},
then
P(e1) + P(e2) + … + P(en) = 1
5-69
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Summary: Rules of Probability
3.
If E and F are disjoint events, then
P(E or F) = P(E) + P(F).
If E and F are not disjoint events, then
P(E or F) = P(E) + P(F) – P(E and F).
5-70
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Summary: Rules of Probability
4.
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If E represents any event and EC
represents the complement of E, then
P(EC) = 1 – P(E).
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Summary: Rules of Probability
5.
If E and F are independent events,
then
P E and F   P E  P F 
5-72
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Section
5.4
Conditional
Probability and
the
General
Multiplication
Rule
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Compute conditional probabilities
2. Compute probabilities using the General
Multiplication Rule
5-74
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Objective 1
• Compute Conditional Probabilities
5-75
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Conditional Probability
The notation P(F|E) is read “the
probability of event F given event E”. It
is the probability that the event F occurs
given that event E has occurred.
5-76
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EXAMPLE
An Introduction to Conditional Probability
Suppose that a single six-sided die is rolled. What is the
probability that the die comes up 4? Now suppose that
the die is rolled a second time, but we are told the
outcome will be an even number. What is the
probability that the die comes up 4?
1
First roll:
S = {1, 2, 3, 4, 5, 6} P(S) 
6
Second roll:
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S = {2, 4, 6}
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1
P(S) 
3
Conditional Probability Rule
If E and F are any two events, then
P E and F  N E and F 
P F E 

P E 
N E 
The probability of event F occurring, given the occurrence
of event E, is found by dividing the probability of E and F
by the probability of E, or by dividing the number of
outcomes in E and F by the number of outcomes in E.
5-78
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EXAMPLE
Conditional Probabilities on Belief about God and Region of the Country
A survey was conducted by the Gallup Organization
conducted May 8 – 11, 2008 in which 1,017 adult
Americans were asked, “Which of the following
statements comes closest to your belief about God – you
believe in God, you don’t believe in God, but you do
believe in a universal spirit or higher power, or you don’t
believe in either?” The results of the survey, by region of
the country, are given in the table on the next slide.
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EXAMPLE
East
Midwest
South
West
Conditional Probabilities on Belief about God and Region of the Country
Believe in
God
204
212
219
152
Believe in
Don’t believe
universal spirit
in either
36
15
29
13
26
9
76
26
(a) What is the probability that a randomly selected adult
American who lives in the East believes in God?
(b) What is the probability that a randomly selected adult
American who believes in God lives in the East?
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EXAMPLE
East
Midwest
South
West
Conditional Probabilities on Belief about God and Region of the Country
Believe in
God
204
212
219
152
Believe in
Don’t believe
universal spirit
in either
36
15
29
13
26
9
76
26
P believes in God lives in the east 
N believe in God and live in the east 

N live in the east 
204

 0.8
204  36  15
5-81
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EXAMPLE
East
Midwest
South
West
Conditional Probabilities on Belief about God and Region of the Country
Believe in
God
204
212
219
152
Believe in
Don’t believe
universal spirit
in either
36
15
29
13
26
9
76
26
P lives in the east believes in God 
N believe in God and live in the east 

N believes in God 
204

 0.26
204  212  219  152
5-82
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EXAMPLE
Murder Victims
In 2005, 19.1% of all murder victims were between
the ages of 20 and 24 years old. Also in 2005, 16.6%
of all murder victims were 20 – 24 year old males.
What is the probability that a randomly selected
murder victim in 2005 was male given that the victim
is 20 – 24 years old?
P male and 20  24 
P male 20  24  
P 20  24 
0.166

 0.869109  0.869  86.9%
0.191
5-83
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Objective 2
• Compute Probabilities Using the General
Multiplication Rule
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General Multiplication Rule
The probability that two events E and F both
occur (E and F are dependent events) is
P E and F   P E  P F E 
In words, the probability of E and F is the
probability of event E occurring times the
probability of event F occurring, given the
occurrence of event E.
5-85
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EXAMPLE
Murder Victims
In 2005, 19.1% of all murder victims were between
the ages of 20 and 24 years old. Also in 2005, 86.9%
of murder victims were male given that the victim was
20 – 24 years old. What is the probability that a
randomly selected murder victim in 2005 was a 20 –
24 year old male?
P male and 20  24   P 20  24  P male 20  24 
 0.869  0.191  0.165979  0.166
5-86
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Section
5.5
Counting
Techniques
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Solve counting problems using the
Multiplication rule
2. Solve counting problems using permutations
3. Solve counting problems using combinations
4. Solve counting problems involving
permutations with nondistinct items
5. Compute probabilities involving
permutations and combinations
5-88
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Objective 1
• Solve Counting Problems Using the
Multiplication Rule
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Multiplication Rule of Counting
If a task consists of a sequence of choices in
which there are p selections for the first choice,
q selections for the second choice, r selections
for the third choice, and so on, then the task of
making these selections can be done in
p  q  r  
different ways.
5-90
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EXAMPLE
Counting the Number of Possible Meals
For each choice of appetizer, we have 4 choices of
entrée, and for each of these 2 • 4 = 8 parings, there
are 2 choices for dessert. A total of
2 • 4 • 2 = 16
different meals can be ordered.
5-91
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If n ≥ 0 is an integer, the factorial
symbol, n!, is defined as follows:
n! = n(n – 1) • • • • • 3 • 2 • 1
0! = 1
1! = 1
5-92
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Objective 2
• Solve Counting Problems Using Permutations
5-93
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A permutation is an ordered
arrangement in which r objects are
chosen from n distinct (different) objects
and repetition is not allowed. The
symbol nPr represents the number of
permutations of r objects selected from
n objects.(order matters)
5-94
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Number of permutations of n Distinct
Objects Taken r at a Time
The number of arrangements of r objects chosen
from n objects, in which
1. the n objects are distinct,
2. repetition of objects is not allowed, and
3. order is important, is given by the formula
n!
n Pr 
n  r !
5-95
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EXAMPLE
Betting on the Trifecta
In how many ways can horses in a 10-horse race
finish first, second, and third?
The 10 horses are distinct. Once a horse crosses the
finish line, that horse will not cross the finish line
again, and, in a race, order is important. We have a
permutation of 10 objects taken 3 at a time.
The top three horses can finish a 10-horse race in
10!
10! 10  9  8  7!


 10  9  8  720 ways
10 P3 
7!
10  3! 7!
5-96
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Objective 3
• Solve Counting Problems Using Combinations
5-97
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A combination is a collection, without
regard to order, of n distinct objects
without repetition. The symbol nCr
represents the number of combinations
of n distinct objects taken r at a time.
5-98
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Number of Combinations of n Distinct
Objects Taken r at a Time
The number of different arrangements of r
objects chosen from n objects, in which
1. the n objects are distinct,
2. repetition of objects is not allowed, and
3. order is not important, is given by the formula
n!
n Cr 
r!n  r !
5-99
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EXAMPLE
Simple Random Samples
How many different simple random samples of size 4
can be obtained from a population whose size is 20?
The 20 individuals in the population are distinct. In
addition, the order in which individuals are selected is
unimportant. Thus, the number of simple random
samples of size 4 from a population of size 20 is a
combination of 20 objects taken 4 at a time.
Use Formula (2) with n = 20 and r = 4:
20!
20! 20 19 18 17 16! 116, 280



 4, 845
20 C 4 
4!20  4 ! 4!16!
4  3 2 116!
24
There are 4,845 different simple random samples of
size 4 from a population whose size is 20.
5-100
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Objective 4
• Solve Counting Problems Involving
Permutations with Nondistinct Items
5-101
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Permutations with Nondistinct Items
The number of permutations of n objects of
which n1 are of one kind, n2 are of a second
kind, . . . , and nk are of a kth kind is given
by
n!
n1 ! n2 !  L  nk !
where n = n1 + n2 + … + nk.
5-102
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EXAMPLE
Arranging Flags
How many different vertical arrangements are there
of 10 flags if 5 are white, 3 are blue, and 2 are red?
We seek the number of permutations of 10 objects,
of which 5 are of one kind (white), 3 are of a second
kind (blue), and 2 are of a third kind (red).
Using Formula (3), we find that there are
10!
10  9  8  7  6  5!

 2,520 different
5!  3!  2!
5!  3!  2!
vertical
arrangements
5-103
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Objective 5
• Compute Probabilities Involving Permutations
and Combinations
5-104
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EXAMPLE
Winning the Lottery
In the Illinois Lottery, an urn contains balls
numbered 1 to 52. From this urn, six balls are
randomly chosen without replacement. For a $1 bet,
a player chooses two sets of six numbers. To win, all
six numbers must match those chosen from the urn.
The order in which the balls are selected does not
matter. What is the probability of winning the
lottery?
5-105
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EXAMPLE
Winning the Lottery
The probability of winning is given by the number of
ways a ticket could win divided by the size of the
sample space. Each ticket has two sets of six
numbers, so there are two chances of winning for
each ticket. The sample space S is the number of
ways that 6 objects can be selected from 52 objects
without replacement and without regard to order, so
N(S) = 52C6.
5-106
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EXAMPLE
Winning the Lottery
The size of the sample space is
52!
N S   52 C6 
6! 52  6 !
52  51 50  49  48  47  46!

 20, 358,520
6!  46!
Each ticket has two sets of 6 numbers, so a player
has two chances of winning for each $1. If E is the
event “winning ticket,” then
2
P E  
 0.000000098
20, 358,520
There is about a 1 in 10,000,000 chance of winning
the Illinois Lottery!
5-107
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Section
5.6
Putting It
Together: Which
Method Do I Use?
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Objectives
1. Determine the appropriate probability rule to
use
2. Determine the appropriate counting
technique to use
5-109
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Objective 1
• Determine the Appropriate Probability Rule to
Use
5-110
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5-111
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5-112
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5-113
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EXAMPLE
Probability: Which Rule Do I Use?
In the game show Deal or No Deal?, a contestant is
presented with 26 suitcases that contain amounts
ranging from $0.01 to $1,000,000. The contestant
must pick an initial case that is set aside as the game
progresses. The amounts are randomly distributed
among the suitcases prior to the game. Consider the
following breakdown:
5-114
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EXAMPLE
Probability: Which Rule Do I Use?
The probability of this event is not compound. Decide
among the empirical, classical, or subjective approaches.
Each prize amount is randomly assigned to one of the 26
suitcases, so the outcomes are equally likely. From the
table we see that 7 of the cases contain at least $100,000.
Letting E = “worth at least $100,000,” we compute P(E)
using the classical approach.
N E 
P E  
N S 
7

 0.269
26
5-115
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EXAMPLE
Probability: Which Rule Do I Use?
N E  7
P E  

 0.269
N S  26
The chance the contestant selects a suitcase worth at
least $100,000 is 26.9%. In 100 different games, we
would expect about 27 games to result in a contestant
choosing a suitcase worth at least $100,000.
5-116
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EXAMPLE
Probability: Which Rule Do I Use?
According to a Harris poll in January 2008, 14% of
adult Americans have one or more tattoos, 50% have
pierced ears, and 65% of those with one or more
tattoos also have pierced ears. What is the probability
that a randomly selected adult American has one or
more tattoos and pierced ears?
P one or more tatoos and pierced ears   P E  P F | E 
 0.14 0.65 
 0.091
5-117
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EXAMPLE
Probability: Which Rule Do I Use?
The probability of a compound event involving ‘AND’.
Letting E = “one or more tattoos” and F = “ears
pierced,” we are asked to find P(E and F). The problem
statement tells us that P(F) = 0.50 and P(F|E) = 0.65.
Because P(F) ≠ P(F|E), the two events are not
independent. We can find P(E and F) using the General
Multiplication Rule.
P E and F   P E  P F | E   0.14 0.65   0.091
So, the chance of selecting an adult American at random
who has one or more tattoos and pierced ears is 9.1%.
5-118
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Objective 2
• Determine the appropriate counting technique
to use
5-119
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5-120
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EXAMPLE
Counting: Which Technique Do I Use?
The Hazelwood city council consists of 5 men and 3
women. How many different subcommittees can be
formed that consist of 3 men and 2 women?
Sequence of events to consider: select the men, then
select the women. Since the number of choices at each
stage is independent of previous choices, we use the
Multiplication Rule of Counting to obtain
N(subcommittees)
= N(ways to pick 3 men) • N(ways to pick 2 women)
5-121
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EXAMPLE
Counting: Which Technique Do I Use?
To select the men, we must consider the number of
arrangements of 5 men taken 3 at a time. Since the
order of selection does not matter, we use the
combination formula.
5!
N ways to pick 3 men   5 C3 
 10
3! 2!
To select the women, we must consider the number of
arrangements of 3 women taken 2 at a time. Since the
order of selection does not matter, we use the
combination formula again.
3!
N ways to pick 3 women   3 C2 
3
2!1!
N(subcommittees) = 10 • 3 = 30. There are 30 possible
subcommittees that contain 3 men and 2 women.
5-122
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EXAMPLE
Counting: Which Technique Do I Use?
On February 17, 2008, the Daytona International
Speedway hosted the 50th running of the Daytona
500. Touted by many to be the most anticipated event
in racing history, the race carried a record purse of
almost $18.7 million. With 43 drivers in the race, in
how many different ways could the top four finishers
(1st, 2nd, 3rd, and 4th place) occur?
5-123
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EXAMPLE
Counting: Which Technique Do I Use?
The number of choices at each stage is independent of
previous choices, so we can use the Multiplication Rule
of Counting. The number of ways the top four finishers
can occur is
N(top four) = 43 • 42 • 41 • 40 = 2,961,840
We could also approach this problem as an arrangement
of units. Since each race position is distinguishable,
order matters in the arrangements. We are arranging the
43 drivers taken 4 at a time, so we are only considering
a subset of r = 4 distinct drivers in each arrangement.
Using our permutation formula, we get
43!
N top four   43 P4 
 43 42  41 40  2,961,840
43  4 !
5-124
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