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Chapter 5
Probability
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Section 5.1 Probability Rules
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5-2
Probability is a measure of the likelihood of a random
phenomenon or chance behavior. Probability describes
the long-term proportion with which a certain outcome
will occur in situations with short-term uncertainty.
Use the probability applet to simulate flipping a coin
100 times. Plot the proportion of heads against the
number of flips. Repeat the simulation.
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5-3
Probability deals with experiments that yield random
short-term results or outcomes, yet reveal long-term
predictability.
The long-term proportion with which a certain
outcome is observed is the probability of that
outcome.
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5-4
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.
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5-5
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 even 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.
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5-6
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)}
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5-7
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5-8
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.
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5-9
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.
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5-11
True or false. The following represents a
probability model.
Cell Phone
Provider
AT&T
Sprint
T–Mobile
Verizon
Probability
0.271
0.236
0.111
0.263
A. True
B. False
Slide 5- 12
Copyright © 2010 Pearson Education, Inc.
True or false. The following represents a
probability model.
Cell Phone
Provider
AT&T
Sprint
T–Mobile
Verizon
Probability
0.271
0.236
0.111
0.263
A. True
B. False
Slide 5- 13
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5-14
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5-15
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.
(Source: http://www.members.tripod.com/~passpigs/prob.html)
Outcome
Frequency
Side with no dot
1344
Side with dot
1294
Razorback
767
Trotter
365
Snouter
137
Leaning Jowler
32
(a) Use the results of the experiment to build a probability model for the way the pig
lands.
(b) Estimate the probability that a thrown pig lands on the “side with dot”.
(c) Would it be unusual to throw a “Leaning Jowler”?
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(a) Outcome
Side with no dot
Probability
1344
 0.341
3939
Side with dot
0.329
Razorback
0.195
Trotter
0.093
Snouter
0.035
Leaning Jowler
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”.
(c) A “Leaning Jowler” would be unusual. We would expect in 1000 throws of the
pig to obtain “Leaning Jowler” about 8 times.
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5-17
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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.
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5-19
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.
(a) There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30 candies, so N(S) = 30.
N (yellow)
N (S )
6

30
 0.2
P(yellow) 
(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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5-20
A box contains 6 twenty-five watt light
bulbs, 9 sixty-watt light bulbs, and 5
hundred-watt light bulbs. What is the
probability a randomly selected light bulb
is sixty-watts?
A. 0.45
B. 0.3
C. 0.05
D. 0.25
Slide 5- 21
Copyright © 2010 Pearson Education, Inc.
A box contains 6 twenty-five watt light
bulbs, 9 sixty-watt light bulbs, and 5
hundred-watt light bulbs. What is the
probability a randomly selected light bulb
is sixty-watts?
A. 0.45
B. 0.3
C. 0.05
D. 0.25
Slide 5- 22
Copyright © 2010 Pearson Education, Inc.
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5-23
EXAMPLE
Using Simulation
Use the probability applet on your calculator (instructor
will show you how) 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.
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5-24
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5-25
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.
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5-26
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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5-27
Section 5.2 Probability Rules
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5-28
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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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5-31
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5-32
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.
Number of Rooms
in Housing Unit
Probability
One
0.010
Two
0.032
Three
0.093
Four
0.176
Five
0.219
Six
0.189
All probabilities are between 0
and 1, inclusive.
Seven
0.122
Eight
0.079
0.010 + 0.032 + … + 0.080 = 1
Nine or more
0.080
(a) Verify that this is a probability
model.
Source: American Community Survey,
U.S. Census Bureau
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Number of Rooms
in Housing Unit
Probability
(b) What is the probability a
randomly selected housing unit
has two or three rooms?
One
0.010
Two
0.032
Three
0.093
Four
0.176
= P(two) + P(three)
Five
0.219
= 0.032 + 0.093
Six
0.189
= 0.125
Seven
0.122
Eight
0.079
Nine or more
0.080
P(two or three)
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Number of Rooms
in Housing Unit
Probability
(c) What is the probability a
randomly selected housing unit
has one or two or three rooms?
One
0.010
Two
0.032
Three
0.093
Four
0.176
P(one or two or three)
Five
0.219
= P(one) + P(two) + P(three)
Six
0.189
= 0.010 + 0.032 + 0.093
Seven
0.122
Eight
0.079
Nine or more
0.080
= 0.135
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5-35
The data shows the distance employees of a
company travel to work. One of these
employees is randomly selected. Determine
the probability the employee travels between
10 and 29 miles to work.
A. 0.401
B. 0.566
C. 0.334
D. 0.735
Slide 5- 36
Distance (miles)
0–9
10 – 19
20 – 29
30 – 39
40 – 49
Copyright © 2010 Pearson Education, Inc.
Number of employees
124
309
257
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2
The data shows the distance employees of a
company travel to work. One of these
employees is randomly selected. Determine
the probability the employee travels between
10 and 29 miles to work.
A. 0.401
B. 0.566
C. 0.334
D. 0.735
Slide 5- 37
Distance (miles)
0–9
10 – 19
20 – 29
30 – 39
40 – 49
Copyright © 2010 Pearson Education, Inc.
Number of employees
124
309
257
78
2
The table shows the favorite pizza topping
for a sample of students. One of these
students is selected at random. Find the
probability the student is female or prefers
sausage.
Cheese Pepperoni Sausage Total
A. 0.458
B. 0.583
Male
Female
Total
8
2
10
C. 0.125
D. 0.556
Slide 5- 38
Copyright © 2010 Pearson Education, Inc.
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The table shows the favorite pizza topping
for a sample of students. One of these
students is selected at random. Find the
probability the student is female or prefers
sausage.
Cheese Pepperoni Sausage Total
A. 0.458
B. 0.583
Male
Female
Total
8
2
10
C. 0.125
D. 0.556
Slide 5- 39
Copyright © 2010 Pearson Education, Inc.
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5-40
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5-41
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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5-42
N (E)
N (S )
6

36
1

6
P( E ) 
N (F )
N (S )
10

36
5

18
P( F ) 
N ( E and F )
N (S )
3

36
1

12
P( E and F ) 
P( E or F )  P( E )  P( F )  P( E and F )
6 10 3
  
36 36 36
13

36
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5-43
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5-44
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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5-45
Complement Rule
If E represents any event and EC represents the complement
of E, then
P(EC) = 1 – P(E)
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5-46
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
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5-47
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?
There are a total of
24,358 + 39,112 + … + 4,895 = 393,186
residents in Hartford County, CT.
The probability a randomly selected
resident will have a commute time of “90
or more minutes” is
Source: United States Census Bureau
4895
 0.012
393,186
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5-48
(b) Compute the probability that a randomly selected resident of
Hartford County, CT will have a commute time less than 90 minutes.
P(less than 90 minutes) = 1 – P(90 minutes or more)
= 1 – 0.012
= 0.988
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5-49
Section 5.3 Independence and 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.
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5-52
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.
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5-53
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5-54
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5-55
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
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5-56
EXAMPLE
Computing Probabilities of Independent Events
A manufacturer of exercise equipment knows that 10% of their products are
defective. They also know that only 30% of their customers will actually use the
equipment in the first year after it is purchased. If there is a one-year warranty
on the equipment, what proportion of the customers will actually make a valid
warranty claim?
We assume that the defectiveness of the equipment is independent of the use
of the equipment. So,
P  defective and used   P  defective   P  used 
 (0.10)(0.30)
 0.03
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5-57
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5-58
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 four 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
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5-59
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5-60
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 – 0.99186500
= 0.9832
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5-61
Forty-four percent of college students have
engaged in binge drinking. If five college
students are randomly selected, what is the
probability that at least one of the five has
engaged in binge drinking?
A. 0.055
B. 0.216
C. 0.945
D. 0.016
Slide 5- 62
Copyright © 2010 Pearson Education, Inc.
Forty-four percent of college students have
engaged in binge drinking. If five college
students are randomly selected, what is the
probability that at least one of the five has
engaged in binge drinking?
A. 0.055
B. 0.216
C. 0.945
D. 0.016
Slide 5- 63
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5-64
Section 5.4 Conditional Probability and the
General Multiplication Rule
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5-65
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5-66
Conditional Probability
The notation P(F | E) is read “the probability of
event F given event E”. It is the probability of
an event F given the occurrence of the event E.
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5-67
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?
First roll:
Second roll:
S = {1, 2, 3, 4, 5, 6}
S = {2, 4, 6}
P(S ) 
1
6
P(S ) 
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1
3
5-68
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5-69
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 below.
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
(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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5-70
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
(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  

N  believe in God and live in the east 
N  live in the east 
204
204  36  15
 0.8
(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
204  212  219  152  0.26
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5-71
EXAMPLE
Murder Victims
In 2005, 19.1% of all murder victims were between the ages of 20 and 24
years old. Also in 1998, 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 20  24  
P  male and 20  24 
P  20  24 
0.166
0.191
= 0.869

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5-72
The table shows the favorite pizza topping
for a sample of students. What is the
probability that a randomly selected student
who was male preferred pepperoni?
A. 0.333
B. 0.375
C. 0.6
Cheese Pepperoni Sausage Total
Male
Female
Total
8
2
10
D. 0.556
Slide 5- 73
Copyright © 2010 Pearson Education, Inc.
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The table shows the favorite pizza topping
for a sample of students. What is the
probability that a randomly selected student
who was male preferred pepperoni?
A. 0.333
B. 0.375
C. 0.6
Cheese Pepperoni Sausage Total
Male
Female
Total
8
2
10
D. 0.556
Slide 5- 74
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Section 5.5
Counting Techniques
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5-75
One, two, three, we’re…
Counting
76
BASIC COUNTING PRINCIPLES
Counting problems are of the following kind:
“How many different 8-letter passwords are
there?”
“How many possible ways are there to pick
11 soccer players out of a 20-player team?”
Most importantly, counting is the basis for
computing probabilities of discrete events.
“What is the probability of winning the
lottery?”
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BASIC COUNTING PRINCIPLES
The sum rule:
If a task can be done in n1 ways and a second task
in n2 ways, and if these two tasks cannot be done at
the same time, then there are n1 + n2 ways to do
either task.
Example:
The department will award a free computer to
either a student or a teacher.
How many different choices are there, if there are
530 students and 45 teachers?
There are 530 + 45 = 575 choices.
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BASIC COUNTING PRINCIPLES
Generalized sum rule:
If we have tasks T1, T2, …, Tm that can
be done in n1, n2, …, nm ways,
respectively, and no two of these tasks
can be done at the same time, then
there are n1 + n2 + … + nm ways to do
one of these tasks.
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BASIC COUNTING PRINCIPLES
The product rule:
Suppose that a procedure can be broken
down into two successive tasks. If there
are n1 ways to do the first task and n2
ways to do the second task after the first
task has been done, then there are n1n2
ways to do the procedure.
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BASIC COUNTING PRINCIPLES
Example: How many different license
plates are there that contain exactly three
English letters ?
Solution: There are 26 possibilities to pick
the first letter, then 26 possibilities for the
second one, and 26 for the last one.
So there are 262626 = 17576 different
license plates.
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BASIC COUNTING PRINCIPLES
Generalized product rule:
If we have a procedure consisting of
sequential tasks T1, T2, …, Tm that can
be done in n1, n2, …, nm ways,
respectively, then there are n1 ⋅ n2 ⋅ … ⋅
nm ways to carry out the procedure.
82
Tree Diagrams
How many bit strings of length
four do NOT
0
0
have two consecutive 1s?
1
0
0
1
1
0
1
1
0
0
1
1
0
1
There are 8 strings.
83
0
1
0
1
0
1
0
1
0
1
0
1
0
1
THE PIGEONHOLE PRINCIPLE
The pigeonhole principle: If (k + 1) or more
objects are placed into k boxes, then there is at
least one box containing two or more of the
objects.
Example 1: If there are 11 players in a soccer
team that wins 12-0, there must be at least one
player in the team who scored at least twice.
Example 2: If you have 6 classes from Monday
to Friday, there must be at least one day on
which you have at least two classes.
84
THE PIGEONHOLE PRINCIPLE
The generalized pigeonhole principle: If N
objects are placed into k boxes, then
there is at least one box containing at
least N/k of the objects.
Example 1: In a 60-student class, at least
12 students will get the same letter grade
(A, B, C, D, or F).
85
THE PIGEONHOLE PRINCIPLE
Example 2: Assume you have a drawer
containing a random distribution of a dozen
brown socks and a dozen black socks. It is
dark, so how many socks do you have to pick
to be sure that among them there is a matching
pair?
Solution: There are two types of socks, so if
you pick at least 3 socks, there must be either
at least two brown socks or at least two black
socks.
86
How many 4-letter television call signs are
possible, if each sign must start with either
a K or a W?
A. 35,152
B. 456,976
C. 16
D. 104
Slide 5- 87
Copyright © 2010 Pearson Education, Inc.
How many 4-letter television call signs are
possible, if each sign must start with either
a K or a W?
A. 35,152
B. 456,976
C. 16
D. 104
Slide 5- 88
Copyright © 2010 Pearson Education, Inc.
There are 15 dogs entered in a show. How
many ways can first, second, and third place
be awarded?
A. 45
B. 455
C. 2,730
D. 3,375
Slide 5- 89
Copyright © 2010 Pearson Education, Inc.
There are 15 dogs entered in a show. How
many ways can first, second, and third place
be awarded?
A. 45
B. 455
C. 2,730
D. 3,375
Slide 5- 90
Copyright © 2010 Pearson Education, Inc.
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reserved
5-91
Permutations and Combinations
How many ways are there to pick a set of 3 people from a group of 6?
The answer to this depends on whether we want the order in which they are
picked to matter or not.
For example, picking person C, then person A, and then person E leads to
the same group as first picking E, then C, and then A.
There are 6 choices for the first person, 5 for the second one, and 4 for the
third one, so there are 6⋅5⋅4 = 120 ways to do this.
Since in the original statement, it does not seem that order is important.
This is not the correct result!
However, these cases are counted separately in the above equation.
92
Permutations and Combinations
So how can we compute how many different subsets
of people can be picked (that is, we want to disregard
the order of picking) ?
To find out about this, we need to first look at
permutations.
A permutation of a set of distinct objects is an ordered
arrangement of these objects.
An ordered arrangement of r elements of a set is
called an r-permutation.
93
Permutations and Combinations
Example: Let S = {1, 2, 3}.
The arrangement 3, 1, 2 is a permutation of S.
The arrangement 3, 2 is a 2-permutation of S.
The number of r-permutations of a set with n
distinct elements is denoted by P(n, r) or nPr .
We can calculate P(n, n) with the product rule:
P(n, n) = n⋅(n – 1)⋅(n – 2) ⋅…⋅3⋅2⋅1.
(n choices for the first element, (n – 1) for the
second one, (n – 2) for the third one…)
94
Permutations and Combinations
Example: 8P3 = (8⋅7⋅6⋅5⋅4⋅3⋅2⋅1)/(5⋅4⋅3⋅2⋅1) =
8⋅7⋅6 = 336
General formula: P(n, r) = n!/(n – r)! = nPr
Knowing this, we can return to our initial
question:
How many ways are there to pick a set of 3
people from a group of 6 (disregarding the
order of picking)?
95
A combination is an arrangement,
without regard to order, of n distinct
objects without repetitions. The symbol
nCr represents the number of
combinations of n distinct objects taken
r at a time, where r < n.
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5-96
PERMUTATIONS AND COMBINATIONS
An r-combination of elements of a set is an unordered
selection of r elements from the set.
Thus, an r-combination is simply a subset of the set
with r elements.
Example: Let S = {1, 2, 3, 4}.
Then {1, 3, 4} is a 3-combination from S.
The number of r-combinations of a set with n distinct
elements is denoted by C(n, r) or nCr.
Example: C(4, 2) = 6, since, for example, the 2combinations of a set {1, 2, 3, 4} are {1, 2}, {1, 3}, {1,
4}, {2, 3}, {2, 4}, {3, 4}.
97
PERMUTATIONS AND COMBINATIONS
How can we calculate C(n, r)?
Consider that we can obtain the r-permutation of a set
in the following way:
First, we form all the r-combinations of the set
(there are C(n, r) such r-combinations).
Then, we generate all possible orderings in each of
these r-combinations (there are P(r, r) such orderings
in each case).
Therefore, we have:
P(n, r) = C(n, r)⋅P(r, r)
98
PERMUTATIONS AND COMBINATIONS
C(n, r) = nCr = P(n, r)/P(r, r) = n!/(n – r)!/(r!/(r –
r)!) = n!/(r!(n – r)!)
Now we can answer our initial question:
How many ways are there to choose a set of 3
people from a group of 6 (disregarding the
order of picking)?
C(6, 3) = 6!/(3!⋅3!) = 720/(6⋅6) = 720/36 = 20
There are 20 different ways, that is, 20 different
groups to be picked.
99
PERMUTATIONS AND COMBINATIONS
Corollary:
Let n and r be nonnegative integers with r ≤ n.
Then C(n, r) = C(n, n – r).
Note that “choosing a group of r people from a
group of n people” is the same as “splitting a
group of n people into a group of r people and
another group of (n – r) people”.
100
PERMUTATIONS AND COMBINATIONS
Example:
A soccer club has 8 female and 7 male
members. For today’s match, the coach wants
to have 6 female and 5 male players on the
grass. How many possible configurations are
there?
8C6
⋅ 7C5
= 28⋅21
= 588
101
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5-102
Determine the value of 9C3.
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5-103
The United States Senate consists of 100
members. In how many ways can 4
members be randomly selected to attend a
luncheon at the White House?
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reserved
5-104
There are 13 students in a club. How many
ways can four students be selected to attend
a conference?
A. 17,160
B. 52
C. 28,561
D. 715
Slide 5- 105
Copyright © 2010 Pearson Education, Inc.
There are 13 students in a club. How many
ways can four students be selected to attend
a conference?
A. 17,160
B. 52
C. 28,561
D. 715
Slide 5- 106
Copyright © 2010 Pearson Education, Inc.
Section 5.6
Bayes’s Rule
The material in this section is available on the CD that accompanies the text.
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5-107
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5-108
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reserved
5-109
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?
● 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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5-110
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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5-111
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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5-112
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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5-113
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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5-114
• 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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5-115
● 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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5-119
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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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5-121
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5-122
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)
● 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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5-125
• 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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5-126
• 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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5-127
● 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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5-128
• 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’ 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( U2 )  P( B | U2 )
• This rule is very useful when P(U1|B) is
difficult to compute, but P(B|U1) is easier
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5-130
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5-131
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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reserved
5-132
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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5-133
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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5-134
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( Ai | E ) 
P( A1 )  P( E | A1 )
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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5-135
EXAMPLE
Bayes’s Rule (continued)
P ( A1 )  P ( E | A1 )
P ( A1 )  P ( E | A1 )  P ( A2 )  P ( E | A2 )



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reserved
.55  .15
.55  .15  .45  .20
.0825
.0825  .0900
.4783
5-136