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Chapter
5
Probability
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Section
5.1
Probability Rules
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Probability is a measure of the likelihood of a
random phenomenon or chance behavior
occurring. 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.
5-3
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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.
5-4
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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.
5-5
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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.
5-6
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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)}
5-7
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EXAMPLE
Identifying Events and the Sample Space of a
Probability Experiment
A probability experiment consists of rolling a single fair
die.
(a) Identify the outcomes of the probability experiment.
(b) Determine the sample space.
(c) Define the event E = “roll an even number
(a) e1 = 1, e2 = 2, e3 = 3, e4 = 4, 𝑒5 = 5, 𝑒6 = 6
(b){1, 2, 3, 4, 5, 6}
(c) {2, 4, 6}
5-8
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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.The sum of the probabilities of all
outcomes must equal 1.
That is, if the sample space
S = {e1, e2, …, en}, then
P(e1) + P(e2) + … + P(en) = 1
5-9
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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.
5-10
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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.
(usually when probability is less than
5%)
5-12
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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
5-13
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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”?
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(a)
Outcome
Side with no dot
5-16
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”.
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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.
5-18
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EXAMPLE
Building a Probability Model
Means of Travel Frequency
The table represents the
153
results in which 200 people Drive alone
Carpool
22
were asked their means of
Public Trans.
10
travel to work.
a) Use the survey data to
Walk
5
build a probability
Other means
3
model for means of
Work at home
7
travel to work.
b) Estimate the probability that a randomly selected
individual carpools to work. Interpret this result.
c) Would it be unusual to randomly select an individual
who walks to work?
5-19
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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.
5-20
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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
5-21
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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.
5-22
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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.
5-23
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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.
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EXAMPLE
Computing Probabilities Using the Classical Method
A pair of fair dice is rolled.
(a) Compute the probability of rolling a seven.
(b) Compute the probability of rolling “snake eyes.”
(c) Comment on the likelihood of rolling a seven versus
rolling a two.
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EXAMPLE
Computing Probabilities Using the Classical Method
Sophia has three tickets to a concert, but Yolanda,
Michael, Kevin, and Marissa all want to go to the
concert with her. To be fair, Sophia randomly selects
the two people who can go with her.
(a) Determine the sample size of the experiment. In
other words, list all possible simple random sample
sizes of n = 2.
(b) Compute the probability of the event “ Michael and
Kevin attend the concert.
(c) Compute the probability of the event “ Marissa
attends the concert.
(d) Interpret the probability of part (c).
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EXAMPLE
Computing Probabilities Using the Classical Method
Suppose that a survey asked 500 families with three
children to disclose the gender of their children and
found that 180 of the families had two boys and one girl.
(a) Estimate the probability of having two boys and one
girl in a three-child family using the empirical method.
(b) Compute and interpret the probability of having two
boys and one girl in a three-child family using the
classical method, assuming boys and girls are equally
likely.
5-27
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EXAMPLE
(a) P(E) =
Computing Probabilities
180
500
= 0.36 so 36%.
(b) Note, S = {BBB, BBG, BGB, BGG, GBB, GBG,
GGB, GGG} so our sample space has a size of 8.
The possibilities of 2 boys and one girl are:
{BBG, BGB, GBB}, hence:
3
8
P(E) = = 0.375,
so 37.5%.
5-28
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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.
5-29
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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.
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Section
5.2
The Addition Rule
and
Complements
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Two events are disjoint if they have no
outcomes in common. Another name for
disjoint events is mutually exclusive
events.
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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.
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5-34
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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 
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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
5-36
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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
Three
Four
Five
0.032
0.093
0.176
0.219
Six
Seven
Eight
0.189
0.122
0.079
Nine or more
0.080
0.010 + 0.032 + … + 0.080 = 1
5-37
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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
5-38
Number of Rooms Probability
in Housing Unit
One
0.010
Two
Three
Four
Five
0.032
0.093
0.176
0.219
Six
Seven
Eight
0.189
0.122
0.079
Nine or more
0.080
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EXAMPLE
The Addition Rule for Disjoint Events
(c) What is the probability
a randomly selected
housing unit has one or
two or three rooms?
P(one or two or three)
= P(one) + P(two) + P(three)
= 0.010 + 0.032 + 0.093
= 0.135
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Number of Rooms Probability
in Housing Unit
One
0.010
Two
Three
Four
Five
0.032
0.093
0.176
0.219
Six
Seven
Eight
0.189
0.122
0.079
Nine or more
0.080
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EXAMPLE
In 1881, Simon Newcomb discovered that digits do not
occur with equal frequency. After studying lots of data,
he assigned probabilities for each of the first digits of
numbers. This is known as Benford’s Law (because a
physicist named Frank Benford proved it at a later date)
and plays a major role in identifying tax fraud.
Digit
1
Probability 0.301
2
3
4
0.176
0.125 0.097
5
6
7
8
9
0.079
0.067
0.058
0.051
0.046
a) Verify that this a probability model.
b) Use Benford’s Law to determine the probability that
a randomly selected first digit is 1 or 2.
c) Use Benford’s Law to determine the probability that
a randomly selected first digit is at least 6.
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EXAMPLE
Suppose that a single card is selected from a standard
52-card deck.
a) Compute the probability of the event E = “drawing
a king.”
b) Compute the probability of the event E = “drawing
a king” or F = “drawing a queen” or G = “drawing a
jack.”
5-41
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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 
5-42
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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.
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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
5-44
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EXAMPLE
Suppose a single card is selected from a standard 52-card
deck. Compute the probability of the event E = “drawing
a king” or F = “drawing a diamond.
5-45
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EXAMPLE
Illustrating the General Addition Rule
Consider the table, which represents the martial status of
males and females 15 years or older in the United States
in 2013.
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Men (in millions)
Female (in millions)
Never married
41.6
36.9
Married
64.4
63.1
Widowed
3.1
11.2
Divorced
11.0
14.4
Separated
2.4
3.2
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EXAMPLE
Illustrating the General Addition Rule
a) Determine the probability that a randomly selected US
resident 15 years or older is male.
b) Determine the probability that a randomly selected US
resident 15 years or older is widowed.
c) Determine the probability that a randomly selected US
resident 15 years or older is widowed or divorced.
d) Determine the probability that a randomly selected US
resident 15 years or older is male or widowed.
5-47
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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.
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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
5-49
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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
5-50
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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
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EXAMPLE
Computing Probabilities Using Complements
There are a total of
24,358 + 39,112 + … + 4,895
= 318,800
residents in Hartford County
The probability a randomly
selected resident will have
a commute time of “90 or
more minutes” is
4895
 0.015
318,800
5-52
Source: United States Census Bureau
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(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 – 0.015
= 0.985
5-53
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EXAMPLE
Illustrating the Complement Rule
The table represents the income distribution of
households in the United States in 2013.
Annual Income
Number (in
thousands)
Annual Income
Number (in
thousands)
Less than $10,000
8940
$50,000 to $74,999
21,659
$10,000 to $14,999
6693
$75,000 to $99,999
14,687
$15,000 to $24,999
13,898
$100,000 to $149,999 15,266
$25,000 to $34,999
12,756
$150,000 to $199,999 6463
$35,000 to $49,999
16,678
$200,000 or more
5-54
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5913
EXAMPLE
Compute the probability that a randomly selected
household earned the following incomes in 2013:
a) $200,000 or more
b) Less than $200,000
c) At least $10,000
5-55
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Section
5.3
Independence
and the
Multiplication
Rule
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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.
5-57
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Disjoint Vs Independent
• Two events are disjoint if they have no
common outcomes, meaning if one event
occurred, the other event could not occur.
• Two events are independent if they do not
affect each others outcome.
• HENCE, two disjoint events are not
independent events.
1-58
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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.
(c) Suppose two 40-year old women live in the same
apartment complex. The events “woman 1 survives the
year” and “woman 2 survives the year” are dependent.
5-59
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EXAMPLE Independent or Not?
(d) Suppose you flip a coin and roll a die. The events
“obtain a tails” and “roll a 5” are independent.
(e) Are the events “earned a bachelor’s degree” and “earn
more than $100,000 per year” independent? No,
because having a bachelor’s affects the likelihood of
earning more than $100,000.
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
5-63
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EXAMPLE
Computing Probabilities of Independent Events
In the game of roulette, the wheel has slots numbered 0,
00, and 1 through 36. A metal ball rolls around a wheel
until it falls into one of the numbered slots. What is the
probability that the ball will land in the slot numbered 17
two times in a row?
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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 
5-65
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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?
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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
5-67
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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
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EXAMPLE
Illustrating the Multiplication Principle for Independent Events
The probability that a randomly selected 24-year-old
male will survive the year is 0.9986 according to the
National Vital Statistics Report.
a) What is the probability that three randomly selected
24-year-old males will survive the year?
b) What is the probability that 20 randomly selected 24year-old males will survive the year?
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Example
Players in sports are said to have “hot streaks” and “cold
streaks.” For example, a batter in baseball might be
considered to be in a slump, or cold strike, if he has
made 10 outs in 10 consecutive at-bats. Suppose that a
hitter successfully reaches base 30% of the time he
comes to the plate.
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Example
a) Find and interpret the probability that the hitter
makes 10 outs in 10 consecutive at-bats assuming
that at-bats are independent events. .7^10
b) Are cold streaks unusual?
YES
c) Find the probability the hitter makes five consecutive
outs and then reaches base safely.
.7^5 x .3
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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
5-72
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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).
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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).
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Summary: Rules of Probability
5.
If E and F are independent events,
then
P E and F   P E  P F 
5-75
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Section
5.4
Conditional
Probability and
the
General
Multiplication
Rule
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
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.
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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:
5-78
S = {2, 4, 6}
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
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.
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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.
5-80
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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
76
9
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
76
P believes in God lives in the east 
9
26
N believe in God and live in the east 

N live in the east 
204

 0.8
204  36  15
5-82
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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
76
P lives in the east believes in God 
9
26
N believe in God and live in the east 

N believes in God 
204

 0.26
204  212  219  152
5-83
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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
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EXAMPLE
Consider the table about martial status and gender of
United States residents aged 15 years old or older in 2013.
Males (in millions)
Females (in millions)
Total
Never Married
41.6
36.9
78.5
Married
64.4
63.1
127.5
Widowed
3.1
11.2
14.3
Divorced
11.0
14.4
25.4
Separated
2.4
3.2
5.6
Totals (in
millions)
122.5
128.8
251.3
5-85
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EXAMPLE
a) Compute the probability that a randomly selected
individual has never married given that the individual
is male.
b) Compute the probability that a randomly selected
individual is male given the individual has never
married.
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EXAMPLE
Suppose that 12.7% of all births are preterm. Also
0.22% of all births resulted in a preterm baby who
weighed 8 pounds, 13 ounces or more. What is the
probability that a randomly selected baby weighs 8
pounds, 13 ounces or more, given that the baby is
preterm? Is this unusual?
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General Multiplication Rule
The probability that two events E and F both
occur 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.
5-88
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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
5-89
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EXAMPLE
The probability that a driver who is speeding gets
pulled over is 0.8. The probability that a driver gets a
ticket, given that he or she is pulled over, is 0.9. What
is the probability that a randomly selected driver who
is speeding gets pulled over and gets a ticket?
P(E and F) = P(E) x P(F|E)
0.8 x 0.9
0.72
5-90
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EXAMPLE
Suppose that of 100 circuits sent to a manufacturing
plant, 5 are defective. The plan manager receiving the
circuits randomly selects 2 and tests them. If both
circuits work, she will accept the shipment.
Otherwise, the shipment is rejected. What is the
probability that the plan manager discovers at least 1
defective circuit and rejects the shipment?
P(at least one defective)
= 1 – P(none defective)
= 1 – P(1st not defective) x P(2nd not defective| 1st not defective)
95
94
= 1 – ( )x( )
100
99
= 1 – 0.902
= 0.098
5-91
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EXAMPLE
In a study to determine whether preferences for self are more or less
prevalent than preferences for others, researchers first asked individuals to
identify the person who is most valuable and likeable to you, or your
favorite other. Of the 1519 individuals surveyed, 42 had chosen themselves
as their favorite other.
a) Suppose we randomly select 1 of the 1519 individuals surveyed. What
is the probability that he or she chosen himself or herself as their
favorite other?
b)
If two individuals from this group are randomly selected, what is the
probability that both chose themselves as their favorite other?
c)
Compute the probability of randomly selecting two individuals from
this group who selected themselves as their favorite other assuming
independence.
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EXAMPLE
a) P(E) =
42
1519
= 0.02765
b) P (𝐸1 and 𝐸2 )
= P(𝐸1 ) x P(𝐸2 |𝐸1 )
=
42
1519
∙
41
1518
~ 0.0007468
c) P(𝐸1 ) = 𝑃( 𝐸2 ) =
So, P (𝐸1 and 𝐸2 )
42
1519
= P(𝐸1 ) x P(𝐸2 |𝐸1 )
42
1519
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∙
42
1519
~ 0.0007645
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Example
• The local golf store sells an “onion bag” that
contains 35 “experienced” golf balls. Suppose
that the bag contains 20 Titleists, 8 Maxflis,
and 7 Top-Flites.
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a) What is the probability that two randomly
selected golf balls are both Titleists? 0.319
b) What is the probability the first ball selected
is Titleist and the second is Maxfli? 0.134
c) What is the probability that the first ball
selected is Maxfli and the second is Titleist? 0.134
d) What is the probability that one golf ball is a
Titleist and the other is a Maxfli?
20
8
P(1st Titlest) + P(1st Maxifli) = 35 x 34
= .2689
1-95
+
8
35
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20
x 34
5.5
Section
Counting
Techniques
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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.
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EXAMPLE
Counting the Number of Possible Meals
The fixed-price dinner at Mabenka Restaurant
provides the following choices:
Appetizer: soup or salad
Entrée: baked chicken, broiled beef patty, baby beef
liver, or roast beef au jus
Dessert: ice cream or cheesecake
How many different meals can be ordered?
5-98
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EXAMPLE
Counting the Number of Possible Meals
Ordering such a meal requires three separate
decisions. Choose an appetizer. 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.
5-99
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EXAMPLE
The International Air Transport Association (IATA)
assigns three-letter codes to represent airport
locations. For example, the code for Fort
Lauderdale International Airport is FLL. How
many different airport codes are possible?
Since all letters are allowed in each space, we get:
26 * 26 * 26 = 17,576
5-100
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EXAMPLE
Three members from a 14-member committee are to
be randomly selected to serve as chair, vice-chair,
and secretary. The first person selected is the chair,
the second is the vice chair, and the third is the
secretary. How many different committee structures
are there?
Note, once someone becomes a chair, they cannot be the other two, or when
someone becomes the vice-chair, they cannot become the secretary.
HENCE:
14 * 13 * 12 = 2184
5-101
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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
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A permutation is an ordered
arrangement in which r objects are
chosen from n distinct (different) objects
so that r ≤ n and repetition is not
allowed. The symbol nPr represents the
number of permutations of r objects
selected from n objects.
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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 !
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Evaluate using formula:
a. 𝑥7 𝑃5
b. 𝑥5 𝑃5
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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!
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EXAMPLE
Twelve people need to be photographed, but
there are only five chairs. (The rest of the
people will be standing behind and no one
cares how they stand.) How many ways can
you sit the twelve people on the five chairs?
Note we have 5 chairs, but once someone sits down in one chair, they
can’t sit in another chair. HENCE:
𝑥12 𝑃5 = 12 * 11 * 10 * 9 * 8 = 95,040
5-107
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A combination is a collection, without
regard to order, in which r objects are
chosen from n distinct objects with r ≤ n
without repetition. The symbol nCr
represents the number of combinations
of n distinct objects taken r at a time.
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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 !
5-109
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Evaluate using formula:
a. 𝑥4 𝐶1
b. 𝑥6 𝐶4
c. 𝑥6 𝐶2
5-110
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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.
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EXAMPLE
Roger, Ken, Tom, and Jay are going to play golf. They
will randomly select teams of two players each. How
many different team combinations can be made?
Since the order of teams does not matter, we get:
𝑥4 𝐶2 = 6
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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
where n = n1 + n2 + … + nk.
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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
5-114
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Example
• How many distinguishable DNA sequences
can be formed using three As, two Cs, two Gs,
and three Ts?
5-115
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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?
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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.
5-117
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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!
5-118
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EXAMPLE
A shipment of 120 fasteners that contains 4
defective fasteners was sent to a manufacturing
plant. The plan’s quality control manager randomly
selects and inspects 5 fasteners. What is the
probability that exactly 1 of the inspected fasteners
is defective?
The way to choose one of the 4 defective parts is (4 𝐶1).
The way to choose 4 of the 116 normal parts is (116 𝐶4).
The total amount of ways to choose 5 parts is (120 𝐶5).
HENCE:
(4 𝐶1)(116 𝐶4)
= 0.1503
(120 𝐶5 )
5-119
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Example
• Suppose that there are 55 Democrats and 45
Republicans in the U.S. Senate. A committee
of seven senators is to be formed by selecting
members of the Senate randomly.
a) What is the probability that the committee is
(55 𝐶7 )
composed of all Democrats?
(100 𝐶7 )
b) What is the probability that the committee is
composed of all Republicans? ((45𝐶𝐶7) )
100 7
c) What is the probability that the committee is
composed of three Democrats and four
(55 𝐶3 )x (45 𝐶4 )
Republicans?
(100 𝐶7 )
1-120
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Section
5.6
Putting It
Together: Which
Method Do I Use?
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5-124
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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:
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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
5-126
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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.
5-127
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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?
5-128
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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%.
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5-130
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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)
5-131
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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.
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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?
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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 !
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Example
• According to some source, about 17% of all 18to 25-year-olds are current marijuana users.
a) What is the probability that four randomly
selected 18- to 25-year-olds are all marijuana
users?
b) What is the probability that among the four
randomly selected 18- to 25-year-olds at least
one is a marijuana user?
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Example
• In the game Text Twist, six letters are given and the player
must form words of varying lengths using the letters provided.
Suppose the letters in a particular game are ENHSIC.
a) How many different arrangements are possible using all 6
letters? 𝑥6 𝑃6
b) How many different arrangements are possible using only 4
letters? 𝑥6 𝑃4
c) The solution to this game has three six-letter words. To
advance to the next round, the player needs at least one of the
six-letter words. If the player simply guesses, what is the
probability that he or she will get one of the six-letter words
on their first guess of six letters?
3
= 0.0042
𝑥6 𝑃6
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Section
5.7
Bayes’s Rule
(on CD)
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Addition Rule for Disjoint Events
If E and F are disjoint (mutually exclusive)
events, then
P E  F   P E   P F 
In general, if E, F, G, … are disjoint
(mutually exclusive) events, then
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General Multiplication Rule
The probability that two events E and F
both occur is
P (E Ç F) = P (E) × P (F E)
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EXAMPLE
Introduction to the Rule of Total Probability
At a university
55% of the students are female and 45% are male
15% of the female students are business majors
20% of the male students are business majors
What percent of students, overall, are business majors?
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EXAMPLE
Introduction to the Rule of Total Probability
● The percent of the business majors in the university
contributed by females
 55% of the students are female
 15% of those students are business majors
 Thus 15% of 55%, or 0.55 • 0.15 = 0.0825 or
8.25% of the total student body are female
business majors
● Contributed by males
 In the same way, 20% of 45%, or
0.45 • 0.20 = .09 or 9% are male business majors
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EXAMPLE
Introduction to the Rule of Total Probability
● Altogether
 8.25% of the total student body are
female business majors
 9% of the total student body are male
business majors
● So … 17.25% of the total student body are
business majors
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EXAMPLE
Introduction to the Rule of Total Probability
Another way to analyze this problem is to use a
tree diagram
Branch 1 – 55% are Female
Branch 2 – 45% are Male
Female
0.55
0.45
Male
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EXAMPLE
Introduction to the Rule of Total Probability
Branch 3 – 15% of Female is business.
Branch 4 – 85% of Female is not business.
Branch 5 – 20% of Male is business.
Branch 6 – 80% of Male is not business.
Female
0.55
0.45
Male
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0.15
Business
0.85
Not Business
0.20
Business
0.80
Not Business
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EXAMPLE
Introduction to the Rule of Total Probability
Multiply out the branches.
Female
0.55
0.45
Male
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0.55•0.15 = 0.0825
0.15
Business
0.85
Not Business 0.55•0.85 = 0.4875
0.20
Business
0.80
Not Business 0.45•0.80 = 0.3600
0.45•0.20 = 0.0900
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EXAMPLE
Introduction to the Rule of Total Probability
Add the two business branches.
Female
0.55
0.45
Male
0.55•0.15 = 0.0825
0.15
Business
0.85
Not Business 0.55•0.85 = 0.4675
0.20
Business
0.80
Not Business 0.45•0.80 = 0.3600
0.45•0.20 = 0.0900
Total = 0.1725
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• This is an example of the Rule of Total
Probability
P(Bus) = 55% • 15% + 45% • 20%
= P(Female) • P(Bus | Female)
+ P(Male) • P(Bus | Male)
• This rule is useful when the sample space can
be divided into two (or more) disjoint parts
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● A partition of the sample space S are two
non-empty sets A1 and A2 that divide up S
● In other words
 A1 ≠ Ø
 A2 ≠ Ø
 A1 ∩ A2 = Ø (there is no overlap)
 A1 U A2 = S (they cover all of S)
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● Let E be any event in the sample space S
● Because A1 and A2 are disjoint, E ∩ A1 and
E ∩ A2 are also disjoint
● Because A1 and A2 cover all of S, E ∩ A1
and E ∩ A2 cover all of E
● This means that we have divided E into two disjoint
pieces E = (E ∩ A1) U (E ∩ A2)
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● Because E ∩ A1 and E ∩ A2 are disjoint, we
can use the Addition Rule
P(E) = P(E ∩ A1) + P(E ∩ A2)
● We now use the General Multiplication Rule
on each of the P(E ∩ A1) and P(E ∩ A2) terms
P(E) = P(A1) • P(E | A1) + P(A2) • P(E | A2)
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●
●
●
P(E) = P(A1) • P(E | A1) + P(A2) • P(E | A2)
This is the Rule of Total Probability (for a
partition into two sets A1 and A2)
It is useful when we want to compute a
probability (P(E)) but we know only pieces of
it (such as P(E | A1))
The Rule of Total Probability tells us how to
put the probabilities together
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● The general Rule of Total Probability assumes
that we have a partition (the general definition)
of S into n different subsets A1, A2, …, An
 Each subset is non-empty
 None of the subsets overlap
 S is covered completely by the union of the
subsets
● This is like the partition before, just that S is
broken up into many pieces, instead of just two
pieces
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Rule of Total Probability
Let E be an event that is a subset of a
sample space S. Let A1, A2, A3, …, An be
partitions of a sample space S. Then,
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EXAMPLE
The Rule of Total Probability
● In a particular town
 30% of the voters are Republican
 30% of the voters are Democrats
 40% of the voters are independents
● This is a partition of the voters into three sets
 There are no voters that are in two sets (disjoint)
 All voters are in one of the sets (covers all of S)
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EXAMPLE
The Rule of Total Probability
● For a particular issue
 90% of the Republicans favor it
 60% of the Democrats favor it
 70% of the independents favor it
● These are the conditional probabilities
 E = {favor the issue}
 The above probabilities are P(E | political party)
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● The total proportion of votes who favor the issue
0.3 • 0.9 + 0.3 • 0.6 + 0.4 • 0.7 = 0.73
● So 73% of the voters favor this issue
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• In our male / female and business / nonbusiness majors examples before, we used the
rule of total probability to answer the question:
“What percent of students are business
majors?”
• We solved this problem by analyzing male
students and female students separately
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• We could turn this problem around
• We were told the percent of female students
who are business majors
• We could also ask:
“What percent of business majors are female?”
• This is the situation for Bayes’s Rule
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● For this example
 We first choose a random business student
(event E)
 What is the probability that this student is
female? (partition element A1)
● This question is asking for the value of P(A1 | E)
● Before, we were working with P(E | A1) instead
 The probability (15%) that a female student is
a business major
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• The Rule of Total Probability
 Know P(Ai) and P(E | Ai)
 Solve for P(E)
• Bayes’s Rule
 Know P(E) and P(E | Ai)
 Solve for P(Ai | E)
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• Bayes’s Rule, for a partition into two sets U1
and U2, is
P(U1 )  P(B | U1 )
P(U1 | B) 
P(U1 )  P(B | U1 )  P(U 2 )  P(B | U 2 )
• This rule is very useful when P(U1|B) is
difficult to compute, but P(B|U1) is easier
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Bayes’s Rule
Let A1, A2, A3, …, An be a partition of a
sample space S. Then, for any event E for
which P(E) > 0, the probability of event Ai for i
= 1, 2, …, n given the event E, is
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EXAMPLE
Bayes’s Rule
The business majors example from before
0.55•0.15 Business
0.0825
0.55•0.85 Not Business
0.4675
0.45•0.20 Business
0.0900
0.45•0.80 Not Business
0.3600
Female
0.55
0.45
Male
Total = 0.1725
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EXAMPLE
Bayes’s Rule
● If we chose a random business major, what is
the probability that this student is female?
 A1 = Female student
 A2 = Male student
 E = business major
● We want to find P(A1 | E), the probability that
the student is female (A1) given that this is a
business major (E)
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EXAMPLE
Bayes’s Rule (continued)
● Do it in a straight way first
 We know that 8.25% of the students are
female business majors
 We know that 9% of the students are male
business majors
 Choosing a business major at random is
choosing one of the 17.25%
● The probability that this student is female is
8.25% / 17.25% = 47.83%
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EXAMPLE
Bayes’s Rule (continued)
● Now do it using Bayes’s Rule – it’s the same
calculation
● Bayes’s Rule for a partition into two sets (n = 2)
P(A1 )  P(E | A1 )
P(Ai | E) 
P(A1 )  P(E | A1 )  P(A2 )  P(E | A2 )
 P(A1) = .55, P(A2) = .45
 P(E | A1) = .15, P(E | A2) = .20
 We know all of the numbers we need
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EXAMPLE
Bayes’s Rule (continued)
P(A1 )  P(E | A1 )
.55 .15

P(A1 )  P(E | A1 )  P(A2 )  P(E | A2 )
.55 .15  .45 .20
.0825

.0825  .0900
 .4783
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