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Chapter
5
Probability
5.1
5.2
5.3
5.4
5.5
5.6
5.7
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.
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.
e1 = boy, boy, e2 = boy, girl, e3 = girl, boy, e4 = girl, girl
(b) Determine the sample space.
{(boy, boy), (boy, girl), (girl, boy), (girl, girl)}
(c) Define the event E = “have one boy”.
{(boy, girl), (girl, boy)}
5-7
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Objective 1
• Apply Rules of Probabilities
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.
0 ≤ P(E) ≤ 1
P(e1) + P(e2) + … + P(en) = 1
5-8
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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, (the
sum of all probabilities must equal 1) is satisfied.
5-9
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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.
5-10
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Objective 2
• Compute and Interpret Probabilities Using the
Empirical Method
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
5-11
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
Building a Probability Model
Pass the PigsTM is a Milton-Bradley 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
Outcome
Freq Prob
(a) Use the results of the
Side with no dot
1344 0.341
experiment to build a
Side with dot
1294 0.329
probability model for the
Razorback
767 0.195
way the pig lands.
Trotter
365 0.093
(b) Estimate the probability
that a thrown pig lands on
Snouter
137 0.035
the “side with dot”.
Leaning Jowler
32
0.008
0.329
(c) Would it be unusual to throw a “Leaning Jowler”?
5-12
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yes
Objective 3
• Compute and Interpret Probabilities Using the
Classical Method
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.
number of ways E can occur m
P E  

number of possible outcomes n
5-13
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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?
P(yellow) = 6/30 = 1/5
(b) What is the probability that it is blue?
P(blue) = 2/30 = 1/15
5-14
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Objective 4
• Use Simulation to Obtain Data Based on
Probabilities
EXAMPLE
Using Simulation
Use the probability applet to simulate throwing a 6-sided
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.
5-15
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Objective 5
• Recognize and Interpret Subjective Probabilities
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.
5-16
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
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.
5-17
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5.1 Summary
1. Apply the rules of probabilities
2. Compute and interpret probabilities using the
empirical method (based on past data values)
3. Compute and interpret probabilities using the
classical method
4. Use simulation to obtain data based on
probabilities
5. Recognize and interpret subjective probabilities
5-18
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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
5-20
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Objective 1
• Use the Addition Rule for Disjoint Events
Two events are disjoint if they have no outcomes in
common. Another name for disjoint events is
mutually exclusive events.
Addition Rule for Disjoint Events
If E and F are disjoint (or mutually exclusive)
events, then
P E or F   P E   P F 
5-21
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We often draw pictures of events using Venn diagrams. 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.”
Let F represent the event “choose a number greater than or equal to 8.”
These events are disjoint as shown in the figure.
P(E) = 3/10
P(F) = 2/10 = 1/5
P(E or F) = P(E) + P(F) = 5/10 = 1/2
5-22
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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
5-23
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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
Four
Five
Six
0.093
0.176
0.219
0.189
Seven
Eight
0.122
0.079
Nine or more
0.080
0.010 + 0.032 + … + 0.080 = 1
5-24
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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
5-25
Number of Rooms Probability
in Housing Unit
One
0.010
Two
0.032
Three
Four
Five
Six
0.093
0.176
0.219
0.189
Seven
Eight
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?
Number of Rooms Probability
in Housing Unit
One
0.010
Two
0.032
Three
Four
P(one or two or three)
Five
= P(one) + P(two) + P(three)
Six
= 0.010 + 0.032 + 0.093
= 0.135
5-26
0.093
0.176
0.219
0.189
Seven
Eight
0.122
0.079
Nine or more
0.080
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Objective 2
• Use the General Addition Rule
The General Addition Rule
For any two events E and F,
P E or F   P E  P F  P E and F 
5-27
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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”
Let F = “the sum <5”
P(E
or
F)

P(E)

P(F)

P(E
and
F)
Find P(E or F)
6 10 3



36 36 36
13

36
5-28
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Objective 3
• Compute the Probability of an Event Using the
Complement Rule
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.
5-29
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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
5-30
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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
5-31
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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?
4895
 0.012
393,186
(b) Compute the probability that
a randomly selected resident
will have a commute time
less than 90 minutes.
Total = 393,186
P(<90 minutes) = 1 – P(> 90 minutes) = 1 – 0.012 = 0.988
5-32
Source: United States Census Bureau
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
5.2 Summary
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
P E or F   P E   P F 
P E or F   P E  P F  P E and F 
5-33
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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
5-35
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Objective 1
1. Identify Independent Events
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.
5-36
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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.
5-37
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Objective 2
• Use the Multiplication Rule for Independent
Events
Multiplication Rule for Independent Events
If E and F are independent events, then
P E and F   P E  P F 
5-38
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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
5-39
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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 oneyear 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
5-40
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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 
5-41
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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?
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
5-42
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Objective 3
• Compute At-Least One Probabilities
P(at least one) = 1 – P(none)
5-43
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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
5-44
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Summary: Rules of Probability
0 ≤ P(E) ≤ 1
P(e1) + P(e2) + … + P(en) = 1
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).
P(EC) = 1 – P(E).
If E and F are independent events, then
P E and F   P E  P F 
5-45
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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
5-47
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Objective 1
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.
P E and F  N E and F 
P F E 

P E 
N E 
5-48
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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?
1
First roll:
S = {1, 2, 3, 4, 5, 6} P(S) 
6
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?
Second roll:
5-49
S = {2, 4, 6}
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1
P(S) 
3
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.
5-50
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EXAMPLE
Conditional Probabilities on Belief about God and Region of the Country
East
Midwest
South
West
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?
P believes in God lives in the east 
204
N believe in God and live in the east 

 0.8

204  36  15
N live in the east 
5-51
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EXAMPLE
Conditional Probabilities on Belief about God and Region of the Country
Believe in
God
Believe in
universal spirit
Don’t believe
in either
East
204
36
15
Midwest
212
29
13
South
219
26
9
West
152
76
26
(b) What is the probability that a randomly selected adult
American who believes in God lives in the East?
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
5-52
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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
5-53
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Objective 2
• Compute Probabilities Using the General
Multiplication Rule
P E and F   P E  P F E 
5-54
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
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
5-55
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5.4 Summary
1. Compute conditional probabilities
2. Compute probabilities using the General
Multiplication Rule
P E and F  N E and F 
P F E 

P E 
N E 
P E and F   P E  P F E 
5-56
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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
5-58
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Objective 1
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.
5-59
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EXAMPLE
Counting the Number of Possible Meals
You have a choice of 2 appetizers, 4 entrées, and 2
desserts.
How many different meals with 1 appetizer, 1
entrée, and 1 dessert may be ordered?
2 • 4 • 2 = 16
5-60
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Objective 2
A permutation is an ordered arrangement in
which r objects are chosen from n distinct
(different) objects and repetitions are not allowed.
The symbol nPr represents the number of
permutations of r objects selected from n objects.
5-61
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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 !
5-62
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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
5-63
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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!
5-64
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Objective 3
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.
5-65
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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 !
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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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Objective 4
• Solve Counting Problems Involving
Permutations with Nondistinct Items
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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.
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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
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Objective 5
• Compute Probabilities Involving Permutations
and Combinations
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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?
The probability of winning is given by the number of ways a ticket
could win divided by the size of the sample space.
P(win) = 2/(52C6)
P E  
2
 0.000000098
20, 358,520
There is about a 1 in 10,000,000 chance of winning the
Illinois Lottery!
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5.5 Summary
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
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Section
5.6
Putting It
Together: Which
Method Do I Use?
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Determine the appropriate probability rule to
use
2. Determine the appropriate counting
technique to use
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Objective 1
• Determine the Appropriate Probability Rule to
Use
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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:
Find the probability of randomly drawing a suitcase worth at
least $100,000.
Answer: 7/26
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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
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Objective 2
• Determine the appropriate counting technique
to use
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EXAMPLE
Counting: Which Technique Do I Use?
The Hazelwood city council consists of 5 men and 4
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)
= 5C3 • 4C2 = 60
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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?
We can use either the multiplication rule or Permutations
43•42•41•40 or 43P4
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= 2,961,840
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5.6 summary
1. Determine the appropriate probability rule to
use
2. Determine the appropriate counting
technique to use
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Section
5.7
Bayes’s Rule
(on CD)
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Use the Rule of Total Probabilities
2. Use Bayes’s Rule to compute probabilities
The material in this section is available on the CD that accompanies the text.
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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
P E  F  G L

 P E   P F   P G  L
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General Multiplication Rule
The probability that two events E and F
both occur is
P E  F   P E  P F E 
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Objective 1
• Use the Rule of Total Probabilities
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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
Female
0.55
0.45
Male
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EXAMPLE
Introduction to the Rule of Total Probability
0.55•0.15 Business
Female
0.55
0.45
Male
0.55•0.85 Not Business
0.45•0.20 Business
0.45•0.80 Not Business
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EXAMPLE
Introduction to the Rule of Total Probability
Multiply out, and add the two business branches
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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• 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,
P E  
P A1  P E A1  P A2  P E A2 L  P An  P E An 
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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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Objective 2
• Use Bayes’s Rule to Compute Probabilities
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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
P Ai  P E Ai 
P Ai E 

P E 
P Ai  P E Ai 
P A1  P E A1  P A2  P E A2 L  P An  P E An 
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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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