Chapter 3
Probability
Larson/Farber 4th ed
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Chapter Outline
• 3.1 Basic Concepts of Probability
• 3.2 Conditional Probability and the Multiplication
Rule
• 3.3 The Addition Rule
• 3.4 Additional Topics in Probability and Counting
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Section 3.1
Basic Concepts of Probability
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Section 3.1 Objectives
•
•
•
•
Identify the sample space of a probability experiment
Identify simple events
Use the Fundamental Counting Principle
Distinguish among classical probability, empirical
probability, and subjective probability
• Determine the probability of the complement of an
event
• Use a tree diagram and the Fundamental Counting
Principle to find probabilities
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Probability Experiments
Probability experiment
• An action, or trial, through which specific results (counts,
measurements, or responses) are obtained.
Outcome
• The result of a single trial in a probability experiment.
Sample Space
• The set of all possible outcomes of a probability
experiment.
Event
• Consists of one or more outcomes and is a subset of the
sample space.
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Probability Experiments
• Probability experiment: Roll a die
• Outcome: {3}
• Sample space: {1, 2, 3, 4, 5, 6}
• Event: {Die is even}={2, 4, 6}
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Example: Identifying the Sample Space
A probability experiment consists of tossing a coin and
then rolling a six-sided die. Describe the sample space.
Solution:
There are two possible outcomes when tossing a coin:
a head (H) or a tail (T). For each of these, there are six
possible outcomes when rolling a die: 1, 2, 3, 4, 5, or
6. One way to list outcomes for actions occurring in a
sequence is to use a tree diagram.
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Solution: Identifying the Sample Space
Tree diagram:
H1 H2 H3 H4 H5 H6
T1 T2 T3 T4 T5 T6
The sample space has 12 outcomes:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
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Simple Events
Simple event
• An event that consists of a single outcome.
 e.g. “Tossing heads and rolling a 3” {H3}
• An event that consists of more than one outcome is
not a simple event.
 e.g. “Tossing heads and rolling an even number”
{H2, H4, H6}
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Example: Identifying Simple Events
Determine whether the event is simple or not.
• You roll a six-sided die. Event B is rolling at least a 4.
Solution:
Not simple (event B has three outcomes: rolling a 4, a 5,
or a 6)
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Fundamental Counting Principle
Fundamental Counting Principle
• If one event can occur in m ways and a second event
can occur in n ways, the number of ways the two
events can occur in sequence is m*n.
• Can be extended for any number of events occurring
in sequence.
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Example: Fundamental Counting
Principle
You are purchasing a new car. The possible
manufacturers, car sizes, and colors are listed.
Manufacturer: Ford, GM, Honda
Car size: compact, midsize
Color: white (W), red (R), black (B), green (G)
How many different ways can you select one
manufacturer, one car size, and one color? Use a tree
diagram to check your result.
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Solution: Fundamental Counting
Principle
There are three choices of manufacturers, two car sizes,
and four colors.
Using the Fundamental Counting Principle:
3 ∙ 2 ∙ 4 = 24 ways
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Types of Probability
Classical (theoretical) Probability
• Each outcome in a sample space is equally likely.
Number of outcomes in event E
• P( E ) 
Number of outcomes in sample space
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Example: Finding Classical Probabilities
You roll a six-sided die. Find the probability of each
event.
1. Event A: rolling a 3
2. Event B: rolling a 7
3. Event C: rolling a number less than 5
Solution:
Sample space: {1, 2, 3, 4, 5, 6}
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Solution: Finding Classical Probabilities
1. Event A: rolling a 3
Event A = {3}
1
P (rolling a 3)   0.167
6
2. Event B: rolling a 7
0
P(rolling a 7)   0
6
Event B= { } (7 is not in
the sample space)
3. Event C: rolling a number less than 5
Event C = {1, 2, 3, 4}
4
P(rolling a number less than 5)   0.667
6
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Types of Probability
Empirical (statistical) Probability
• Based on observations obtained from probability
experiments.
• Relative frequency of an event.
Frequency of event E f

• P( E ) 
Total frequency
n
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Example: Finding Empirical Probabilities
A company is conducting an online survey of randomly
selected individuals to determine if traffic congestion is
a problem in their community. So far, 320 people have
responded to the survey. What is the probability that the
next person that responds to the survey says that traffic
congestion is a serious problem in their community?
Response
Number of times, f
Serious problem
123
Moderate problem
115
Not a problem
82
Σf = 320
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Solution: Finding Empirical Probabilities
Response
event
Number of times, f
Serious problem
123
Moderate problem
115
Not a problem
82
frequency
Σf = 320
f 123
P( Serious problem)  
 0.384
n 320
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Law of Large Numbers
Law of Large Numbers
• As an experiment is repeated over and over, the
empirical probability of an event approaches the
theoretical (actual) probability of the event.
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Types of Probability
Subjective Probability
• Intuition, educated guesses, and estimates.
• e.g. A doctor may feel a patient has a 90% chance of a
full recovery.
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Example: Classifying Types of Probability
Classify the statement as an example of classical,
empirical, or subjective probability.
1. The probability that you will be married by age
30 is 0.50.
Solution:
Subjective probability (most likely an educated guess)
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Example: Classifying Types of Probability
Classify the statement as an example of classical,
empirical, or subjective probability.
2. The probability that a voter chosen at random will
vote Republican is 0.45.
Solution:
Empirical probability (most likely based on a survey)
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Example: Classifying Types of Probability
Classify the statement as an example of classical,
empirical, or subjective probability.
3. The probability of winning a 1000-ticket raffle with
1
one ticket is 1000.
Solution:
Classical probability (equally likely outcomes)
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Range of Probabilities Rule
Range of probabilities rule
• The probability of an event E is between 0 and 1,
inclusive.
• 0 ≤ P(E) ≤ 1
Impossible
Unlikely
Even
chance
[
0
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Likely
Certain
]
0.5
1
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Complementary Events
Complement of event E
• The set of all outcomes in a sample space that are not
included in event E.
• Denoted E ′ (E prime)
• P(E ′) + P(E) = 1
• P(E) = 1 – P(E ′)
E′
• P(E ′) = 1 – P(E)
E
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Example: Probability of the Complement
of an Event
You survey a sample of 1000 employees at a company
and record the age of each. Find the probability of
randomly choosing an employee who is not between 25
and 34 years old.
Employee ages Frequency, f
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
Σf = 1000
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Solution: Probability of the Complement
of an Event
• Use empirical probability to
find P(age 25 to 34)
f
366
P(age 25 to 34)  
 0.366
n 1000
• Use the complement rule
366
P(age is not 25 to 34)  1 
1000
634

 0.634
1000
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Employee ages
Frequency, f
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
Σf = 1000
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Example: Probability Using a Tree
Diagram
A probability experiment consists of tossing a coin and
spinning the spinner shown. The spinner is equally
likely to land on each number. Use a tree diagram to
find the probability of tossing a tail and spinning an odd
number.
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Solution: Probability Using a Tree
Diagram
Tree Diagram:
H
T
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
H1 H2 H3 H4 H5 H6 H7 H8
T1 T2 T3 T4 T5 T6 T7 T8
4 1
  0.25
P(tossing a tail and spinning an odd number) =
16 4
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Textbook Exercises. Page 140
• Exercises 41 – 44
Tree Diagram
Yellow
{1,2,3,4,5,6}
red
{1,2,3,4,5,6}
Blue
{1,2,3,4,5,6}
green
{1,2,3,4,5,6}
experiment




# 41: P(rolling a 5 and landing on blue) = 1/24
# 42: P(rolling an odd number and landing on green) = 3/24
# 43: P(rolling a number less than 6 and landing on yellow) = 5/24
# 44: P(not rolling a number less than 6 and landing on yellow) = 1/24
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Example: Probability Using the
Fundamental Counting Principle
Your college identification number consists of 8 digits.
Each digit can be 0 through 9 and each digit can be
repeated. What is the probability of getting your college
identification number when randomly generating eight
digits?
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Solution: Probability Using the
Fundamental Counting Principle
• Each digit can be repeated
• There are 10 choices for each of the 8 digits
• Using the Fundamental Counting Principle, there are
10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10
= 108 = 100,000,000 possible identification numbers
• Only one of those numbers corresponds to your ID
number
1
P(your ID number) =
100, 000, 000
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Textbook Exercises. Page 140
•
Exercise 46
a.
Total number of possible codes are 26  9 10 10 5  117,000
first number represents 26 possible letters, second number represents 9 digits
that can be used as zero is not allowed, third and fourth numbers are 10
possible digits where as the last number represents five possible even digits
that can be used.
a.
Probability of selecting the correct access code on the first try is
1
117,000
c.
Probability of not selecting the correct access code on the first try is
complement of part b.
= 1
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1
116,999

117,000 117,000
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Odds
• Odds in favor (of winning) is the ratio of the number of successful
outcomes to the number of unsuccessful outcomes.
• Odds against (of losing) is the ratio of the number of unsuccessful
outcomes to the number of successful outcomes.
• Example: Rolling a single die, what are the odds of obtaining an
even number?
Successful outcomes: {2, 4, 6}
Unsuccessful outcomes: {1, 3, 5}
Odds of obtaining an even number : 3 : 3 or 1: 1
Odds and Probability
• If the odds in favor of an event are a : b, then the probability that
the event will occur is given by
a
a+b
• Example: The odds of survival of a cancer patient are 3:5.
What is the probability the patient will survive?
P(survive) =
3
3

 37.5%
3+5
8
5
5

 62 .5%
P (not survive) =
3+5
8
Textbook Exercises. Page 143
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Problem 72
a. P(stock price is less than $21) = 0.25 (or 25%)
b. P(stock price is between $21 and $50) = 0.50 (or 50%)
c. P(stock price is more than $30) = 0.50 (or 50%)
•
Problem 79
Odds that the card picked is a spade are:
# successful outcomes : # unsuccessful outcomes
that is,
# spades in a deck : # other cards in a deck
13 : 39 OR
1:3
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Section 3.1 Summary
• Identified the sample space of a probability
experiment
• Identified simple events
• Used the Fundamental Counting Principle
• Distinguished among classical probability, empirical
probability, and subjective probability
• Determined the probability of the complement of an
event
• Used a tree diagram and the Fundamental Counting
Principle to find probabilities
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Section 3.2
Conditional Probability and the
Multiplication Rule
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Section 3.2 Objectives
• Determine conditional probabilities
• Distinguish between independent and dependent
events
• Use the Multiplication Rule to find the probability of
two events occurring in sequence
• Use the Multiplication Rule to find conditional
probabilities
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Conditional Probability
Conditional Probability
• The probability of an event occurring, given that
another event has already occurred
• Denoted P(B | A) (read “probability of B, given A”)
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Example: Finding Conditional
Probabilities
Two cards are selected in sequence from a standard
deck. Find the probability that the second card is a
queen, given that the first card is a king. (Assume that
the king is not replaced.)
Solution:
Because the first card is a king and is not replaced, the
remaining deck has 51 cards, 4 of which are queens.
4
P( B | A)  P(2 card is a Queen |1 card is a King ) 
 0.078
51
nd
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Example: Finding Conditional
Probabilities
The table shows the results of a study in which
researchers examined a child’s IQ and the presence of a
specific gene in the child. Find the probability that a
child has a high IQ, given that the child has the gene.
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Gene
Present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
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Solution: Finding Conditional
Probabilities
There are 72 children who have the gene. So, the
sample space consists of these 72 children.
Gene
Present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
Of these, 33 have a high IQ.
P( B | A)  P(high IQ | gene present ) 
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 0.458
72
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Independent and Dependent Events
Independent events
• The occurrence of one of the events does not affect
the probability of the occurrence of the other event
• P(B | A) = P(B) or P(A | B) = P(A)
• Events that are not independent are dependent
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Example: Independent and Dependent
Events
Decide whether the events are independent or dependent.
1. Selecting a king from a standard deck (A), not
replacing it, and then selecting a queen from the deck
(B).
Solution:
P( B | A)  P(2nd card is a Queen |1st card is a King ) 
P ( B )  P (Queen) 
4
52
4
51
Dependent (the occurrence of A changes the probability
of the occurrence of B)
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Example: Independent and Dependent
Events
Decide whether the events are independent or dependent.
2. Tossing a coin and getting a head (A), and then
rolling a six-sided die and obtaining a 6 (B).
Solution:
P( B | A)  P(rolling a 6 | head on coin) 
P( B)  P(rolling a 6) 
1
6
1
6
Independent (the occurrence of A does not change the
probability of the occurrence of B)
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The Multiplication Rule
Multiplication rule for the probability of A and B
• The probability that two events A and B will occur in
sequence is
 P(A and B) = P(A) ∙ P(B | A)
• For independent events the rule can be simplified to
 P(A and B) = P(A) ∙ P(B)
 Can be extended for any number of independent
events
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Example: Using the Multiplication Rule
Two cards are selected, without replacing the first card,
from a standard deck. Find the probability of selecting a
king and then selecting a queen.
Solution:
Because the first card is not replaced, the events are
dependent.
P( K and Q)  P( K )  P(Q | K )
4 4
 
52 51
16

 0.006
2652
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Example: Using the Multiplication Rule
A coin is tossed and a die is rolled. Find the probability
of getting a head and then rolling a 6.
Solution:
The outcome of the coin does not affect the probability
of rolling a 6 on the die. These two events are
independent.
P( H and 6)  P( H )  P(6)
1 1
 
2 6
1
  0.083
12
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Example: Using the Multiplication Rule
The probability that a particular knee surgery is
successful is 0.85. Find the probability that three knee
surgeries are successful.
Solution:
The probability that each knee surgery is successful is
0.85. The chance for success for one surgery is
independent of the chances for the other surgeries.
P(3 surgeries are successful) = (0.85)(0.85)(0.85)
≈ 0.614
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Example: Using the Multiplication Rule
Find the probability that none of the three knee
surgeries is successful.
Solution:
Because the probability of success for one surgery is
0.85. The probability of failure for one surgery is
1 – 0.85 = 0.15
P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15)
≈ 0.003
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Example: Using the Multiplication Rule
Find the probability that at least one of the three knee
surgeries is successful.
Solution:
“At least one” means one or more. The complement to
the event “at least one successful” is the event “none are
successful.” Using the complement rule
P(at least 1 is successful) = 1 – P(none are successful)
≈ 1 – 0.003
= 0.997
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Example: Using the Multiplication Rule to
Find Probabilities
More than 15,000 U.S. medical school seniors applied to
residency programs in 2007. Of those, 93% were matched to
a residency position. Seventy-four percent of the seniors
matched to a residency position were matched to one of their
top two choices. Medical students electronically rank the
residency programs in their order of preference and program
directors across the United States do the same. The term
“match” refers to the process where a student’s preference
list and a program director’s preference list overlap,
resulting in the placement of the student for a residency
position. (Source: National Resident Matching Program)
(continued)
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Example: Using the Multiplication Rule to
Find Probabilities
1. Find the probability that a randomly selected senior was
matched a residency position and it was one of the
senior’s top two choices.
Solution:
A = {matched to residency position}
B = {matched to one of two top choices}
P(A) = 0.93 and P(B | A) = 0.74
P(A and B) = P(A)∙P(B | A) = (0.93)(0.74) ≈ 0.688
dependent events
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Example: Using the Multiplication Rule to
Find Probabilities
2. Find the probability that a randomly selected senior that
was matched to a residency position did not get matched
with one of the senior’s top two choices.
Solution:
Use the complement:
P(B′ | A) = 1 – P(B | A)
= 1 – 0.74 = 0.26
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Textbook Exercises. Pages 152
•
Problem 20
Nursing Majors
a. P (nursing major) =
b. P (male) =
1167
 0.294
3964
1255
 0.317
3964
c. P(nursing major given male) =
151
 0.120
1255
d. P (nursing major and male) = P(nursing major) ∙ P(male given nursing major)
 1167 151 


  0.038
3964
1167



e. Dependent events because P(nursing majors given male)  P (nursing majors)
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Textbook Exercises. Page 153
•
Problem 26
Let A = { 1st battery fails the test} and
B = {2nd battery fails the test}
Note that A and B are dependent events. Since batteries are chosen without
replacement probability of event A may affect the probability of event B and viceversa.
a. P(A and B) = P(A) ∙ P(B│A) = (4/16) ∙ (3/15) = 0.05
b. P(A′ and B′) = P(A′) ∙ P(B′│A′) = (12/16) ∙ (11/15) = 0.55
c. P(at least one bulb failed)
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= 1 – P(none failed)
= 1 – P(A′ and B′)
= 1 – 0.55
= 0.45
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Textbook Exercise. Pages 154
•
Problem 30
Note that all three people are independent as they are unrelated. Thus, this is the
case of independent events.
a. P(all three have A+) = (0.31) ∙ (0.31) ∙ (0.31) = 0.0297
b. P(none have A+) = (0.69) ∙ (0.69) ∙ (0.69) = 0.329
c. P(at least one has A+) = 1 – P(none have A+) = 1 – 0.329 = 0.671
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Section 3.2 Summary
• Determined conditional probabilities
• Distinguished between independent and dependent
events
• Used the Multiplication Rule to find the probability
of two events occurring in sequence
• Used the Multiplication Rule to find conditional
probabilities
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Section 3.3
Addition Rule
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Section 3.3 Objectives
• Determine if two events are mutually exclusive
• Use the Addition Rule to find the probability of two
events
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Mutually Exclusive Events
Mutually exclusive
• Two events A and B cannot occur at the same time
A
B
A and B are mutually
exclusive
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A
B
A and B are not mutually
exclusive
63
Example: Mutually Exclusive Events
Decide if the events are mutually exclusive.
Event A: Roll a 3 on a die.
Event B: Roll a 4 on a die.
Solution:
Mutually exclusive (The first event has one outcome, a
3. The second event also has one outcome, a 4. These
outcomes cannot occur at the same time.)
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Example: Mutually Exclusive Events
Decide if the events are mutually exclusive.
Event A: Randomly select a male student.
Event B: Randomly select a nursing major.
Solution:
Not mutually exclusive (The student can be a male
nursing major.)
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The Addition Rule
Addition rule for the probability of A or B
• The probability that events A or B will occur is
 P(A or B) = P(A) + P(B) – P(A and B)
• For mutually exclusive events A and B, the rule can
be simplified to
 P(A or B) = P(A) + P(B)
 Can be extended to any number of mutually
exclusive events
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Example: Using the Addition Rule
You select a card from a standard deck. Find the
probability that the card is a 4 or an ace.
Solution:
The events are mutually exclusive (if the card is a 4, it
cannot be an ace)
Deck of 52 Cards
P(4 or ace)  P(4) + P(ace)
4
4
 +
52 52
8

 0.154
52
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4♣
4♥
4♠
4♦
A♣
A♠ A♥
A♦
44 other cards
67
Example: Using the Addition Rule
You roll a die. Find the probability of rolling a number
less than 3 or rolling an odd number.
Solution:
The events are not mutually exclusive (1 is an
outcome of both events)
Roll a Die
4
Odd
3
5
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6
Less than
1 three
2
68
Solution: Using the Addition Rule
Roll a Die
4
Odd
3
5
6
Less than
1 three
2
P(less than 3 or odd )
 P(less than 3) + P(odd )  P(less than 3 and odd )
2 3 1 4
 +    0.667
6 6 6 6
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Example: Using the Addition Rule
The frequency distribution shows
the volume of sales (in dollars)
and the number of months a sales
representative reached each sales
level during the past three years.
If this sales pattern continues,
what is the probability that the
sales representative will sell
between $75,000 and $124,999
next month?
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Sales volume ($)
Months
0–24,999
3
25,000–49,999
5
50,000–74,999
6
75,000–99,999
7
100,000–124,999
9
125,000–149,999
2
150,000–174,999
3
175,000–199,999
1
70
Solution: Using the Addition Rule
• A = monthly sales between
$75,000 and $99,999
• B = monthly sales between
$100,000 and $124,999
• A and B are mutually exclusive
P( A or B)  P( A) + P( B)
7 9
 +
36 36
16

 0.444
36
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Sales volume ($)
Months
0–24,999
3
25,000–49,999
5
50,000–74,999
6
75,000–99,999
7
100,000–124,999
9
125,000–149,999
2
150,000–174,999
3
175,000–199,999
1
71
Example: Using the Addition Rule
A blood bank catalogs the types of blood given by
donors during the last five days. A donor is selected at
random. Find the probability the donor has type O or
type A blood.
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
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72
Solution: Using the Addition Rule
The events are mutually exclusive (a donor cannot have
type O blood and type A blood)
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
P(type O or type A)  P(type O) + P(type A)
184 164

+
409 409
348

 0.851
409
Larson/Farber 4th ed
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Example: Using the Addition Rule
Find the probability the donor has type B or is Rhnegative.
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
Solution:
The events are not mutually exclusive (a donor can have
type B blood and be Rh-negative)
Larson/Farber 4th ed
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Solution: Using the Addition Rule
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
P(type B or Rh  neg )
 P(type B) + P( Rh  neg )  P(type B and Rh  neg )
45
65
8
102

+


 0.249
409 409 409 409
Larson/Farber 4th ed
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Textbook Exercises. Pages 162 - 164
•
Problem 14
Conference
a. The two events are not mutually exclusive because they can occur at the same
time. As per the data, it is possible to select a female who is a college Prof.
b. P(F or Coll. Prof)
= P(F) + P(Coll. Prof) – P(F and Coll. Prof)
=
•
2575 2110 960
+

 0.753
4950 4950 4950
Problem 20
Tacoma Narrows Bridge
a. P(two occupants)
= 0.298 or 29.8%
b. P(two or more occupants)
= 1 – P(one occupant) ( complement rule)
= 1 – 0.555 = 0.445
c. P(between 2 and 5) = P(2 or 3 or 4 or 5)
= 0.298 + 0.076 + 0.047 + 0.014 = 0.435
Note that in a selected car you can either have 2 or 3 or 4 or 5 occupants but not
all at the same time. Hence they are all mutually exclusive.
Larson/Farber 4th ed
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Textbook Exercises. Pages 162 - 164
•
Problem 24
Left Handed People
Let A = {left handed} and B = {male}
113
475
50

+

 0.538
a. P (A or B′) = P(A) + P(B′) – P(A and B′)
1000
1000
1000
b. P(A′ or B) = P(A′) + (B) – (A′ and B)

887
525
462
+

 0.950
1000 1000 1000
c. P (A or B) = P(A) + P(B) – P(A and B)

113
525
63
+

 0.575
1000 1000 1000
d. P(right handed woman) = P(A′ and B′)

425
 0.425
1000
e. Not mutually exclusive. As per the data given, a woman can be right handed.
Larson/Farber 4th ed
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Section 3.3 Summary
• Determined if two events are mutually exclusive
• Used the Addition Rule to find the probability of two
events
Larson/Farber 4th ed
78
Section 3.4
Additional Topics in Probability and
Counting
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Section 3.4 Objectives
• Determine the number of ways a group of objects can
be arranged in order
• Determine the number of ways to choose several
objects from a group without regard to order
• Use the counting principles to find probabilities
Larson/Farber 4th ed
80
Permutations
Permutation
• An ordered arrangement of objects
• The number of different permutations of n distinct
objects is n! (n factorial)
 n! = n∙(n – 1)∙(n – 2)∙(n – 3)∙ ∙ ∙3∙2 ∙1
 0! = 1
 Examples:
• 6! = 6∙5∙4∙3∙2∙1 = 720
• 4! = 4∙3∙2∙1 = 24
Larson/Farber 4th ed
81
Example: Permutation of n Objects
The objective of a 9 x 9 Sudoku number
puzzle is to fill the grid so that each
row, each column, and each 3 x 3 grid
contain the digits 1 to 9. How many
different ways can the first row of a
blank 9 x 9 Sudoku grid be filled?
Solution:
The number of permutations is
9!= 9∙8∙7∙6∙5∙4∙3∙2∙1 = 362,880 ways
Larson/Farber 4th ed
82
Permutations
Permutation of n objects taken r at a time
• The number of different permutations of n distinct
objects taken r at a time
n!
■
where r ≤ n
n Pr 
( n  r )!
Larson/Farber 4th ed
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Example: Finding nPr
Find the number of ways of forming three-digit codes in
which no digit is repeated.
Solution:
• You need to select 3 digits from a group of 10
• n = 10, r = 3
10!
10!

10 P3 
(10  3)!
7!
10  9  8  7  6  5  4  3  2  1

7  6  5  4  3  2 1
 720 ways
Larson/Farber 4th ed
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Example: Finding nPr
Forty-three race cars started the 2007 Daytona 500.
How many ways can the cars finish first, second, and
third?
Solution:
• You need to select 3 cars from a group of 43
• n = 43, r = 3
43!
43!

43 P3 
(43  3)! 40!
 43  42  41
 74, 046 ways
Larson/Farber 4th ed
85
Distinguishable Permutations
Distinguishable Permutations
• The number of distinguishable permutations of n
objects where n1 are of one type, n2 are of another
type, and so on
n!
■
n1 ! n2 ! n3 !   nk !
where n1 + n2 + n3 +∙∙∙+ nk = n
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Example: Distinguishable Permutations
A building contractor is planning to develop a
subdivision that consists of 6 one-story houses, 4 twostory houses, and 2 split-level houses. In how many
distinguishable ways can the houses be arranged?
Solution:
• There are 12 houses in the subdivision
• n = 12, n1 = 6, n2 = 4, n3 = 2
12!
6! 4! 2!
 13, 860 distinguishable ways
Larson/Farber 4th ed
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Combinations
Combination of n objects taken r at a time
• A selection of r objects from a group of n objects
without regard to order
n!
■
n Cr 
( n  r )! r !
Larson/Farber 4th ed
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Example: Combinations
A state’s department of transportation plans to develop a
new section of interstate highway and receives 16 bids
for the project. The state plans to hire four of the
bidding companies. How many different combinations
of four companies can be selected from the 16 bidding
companies?
Solution:
• You need to select 4 companies from a group of 16
• n = 16, r = 4
• Order is not important
Larson/Farber 4th ed
89
Solution: Combinations
16!
16 C4 
(16  4)!4!
16!

12!4!
16 15 14 13 12!

12! 4  3  2 1
 1820 different combinations
Larson/Farber 4th ed
90
Example: Finding Probabilities
A student advisory board consists of 17 members. Three
members serve as the board’s chair, secretary, and
webmaster. Each member is equally likely to serve any
of the positions. What is the probability of selecting at
random the three members that hold each position?
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Solution: Finding Probabilities
• There is only one favorable outcome
• There are
17!
17 P3 
(17  3)!
17!

 17 16 15  4080
14!
ways the three positions can be filled
1
P( selecting the 3 members) 
 0.0002
4080
Larson/Farber 4th ed
92
Example: Finding Probabilities
You have 11 letters consisting of one M, four Is, four
Ss, and two Ps. If the letters are randomly arranged in
order, what is the probability that the arrangement spells
the word Mississippi?
Larson/Farber 4th ed
93
Solution: Finding Probabilities
• There is only one favorable outcome
• There are
11!
 34, 650
1! 4! 4! 2!
11 letters with 1,4,4, and 2
like letters
distinguishable permutations of the given letters
1
P( Mississippi) 
 0.000029
34650
Larson/Farber 4th ed
94
Example: Finding Probabilities
A food manufacturer is analyzing a sample of 400 corn
kernels for the presence of a toxin. In this sample, three
kernels have dangerously high levels of the toxin. If
four kernels are randomly selected from the sample,
what is the probability that exactly one kernel contains a
dangerously high level of the toxin?
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Solution: Finding Probabilities
• The possible number of ways of choosing one toxic
kernel out of three toxic kernels is
3C1 = 3
• The possible number of ways of choosing three
nontoxic kernels from 397 nontoxic kernels is
397C3 = 10,349,790
• Using the Multiplication Rule, the number of ways of
choosing one toxic kernel and three nontoxic kernels
is
3C1 ∙ 397C3 = 3 ∙ 10,349,790 3 = 31,049,370
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Solution: Finding Probabilities
• The number of possible ways of choosing 4 kernels
from 400 kernels is
400C4 = 1,050,739,900
• The probability of selecting exactly 1 toxic kernel is
C1  397 C3
P(1 toxic kernel ) 
400 C4
3
31, 049,370

 0.0296
1, 050, 739,900
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Textbook Exercises. Pages 174 - 176
•
Problem 20
Skiing
Since there are no ties the skiers can finish in 8! ways = 40,320
Each one takes one place from first to eighth. Thus by fundamental counting
principle they can finish the race in 8! ways.
Note: On your graphing calculator TI 83/84, first enter 8 then press the MATH key,
move under PRB menu and select item 4. Hit ENTER KEY to see the result.
•
Problem 26
Experimental Group
Since the four subjects are not selected in any particular order, the selection is made
using combination. So total number of possible groups of 4 are
20
C4  4845
Note: On your TI 83/84, first enter 20 then press the MATH key, move under PRB
menu and select item 3, then enter 4. Hit the ENTER key to see the result.
Larson/Farber 4th ed
98
Textbook Exercises. Pages 174 - 176
•
Problem 46
Area code
a. Total number of possible 3 digit area codes, without any restrictions, are
10 ∙ 10 ∙ 10 = 1000 area codes.
(Fundamental Counting Principle)
b. Total number of possible 3 digit area codes, if the first digit can’t be a 0 or 1,
are
8 ∙ 10 ∙ 10 = 800 area codes.
c. In this part we have to find probability of selecting an area code that ends in an
odd number if the first digit cannot be 0 or 1.
Recall
totalnumberof outcomesin theeventE
P(E)
totalnumberof outcomesin thesample space
The sample space is all possible area codes whose first digit is neither 0 nor 1 and
the event is all possible area codes from the above mentioned sample space that
end in an odd number.

Larson/Farber 4th ed
8 10  5
 0 .5
8 10 10
99
Section 3.4 Summary
• Determined the number of ways a group of objects
can be arranged in order
• Determined the number of ways to choose several
objects from a group without regard to order
• Used the counting principles to find probabilities
Larson/Farber 4th ed
100