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Sullivan, 4th ed.
Copyright of the definitions and examples is
reserved to Pearson Education, Inc.. In order to
use this PowerPoint presentation, the required
textbook for the class is the Fundamentals of
Statistics, Informed Decisions Using Data,
Michael Sullivan, III, fourth edition.
Probability
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
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.
If we flip a coin 100 times and compute the
proportion of heads observed after each toss
of the coin, what will the proportion
approach?
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.
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Experiment – any activity that leads to well-defined results called
outcomes.
Outcome – the result of a single trial of probability experiment.
Sample Space, S – the set of all possible outcomes of a probability
experiment.
Event, E – a subset of sample space.
Simple event, ei – an event with one outcome is called a simple
event.
Compound event – consists of two or more outcomes.
Example 1 : A die is tossed one time.
(a) List the elements of the sample space S .
(b) List the elements of the event consisting of a number
that is greater than 4.
(a) S  {1, 2, 3, 4, 5, 6 }
(b) E  { 5, 6 }
Example 2 : A coin is tossed twice. List the elements of the sample
space S , and list the elements of the event consisting of
at least one head.
H
H
S  { HH , HT , TH , TT }
T
H
E  { HT , TH , HH }
T
T
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective B : Requirements for Probabilities
1. Each probability must lie on between 0 and 1. (0  P( E )  1)
2. The sum of probabilities for all simple events in S equals 1.
( P(ei )  1)
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. Typically, an event with a probability less than 0.05 is
considered as unusual.
Probabilities should be expressed as reduced fractions or rounded
to three decimal places.
Example 1 : A probability experiment is conducted. Which of these can
be considered a probability of an outcome?
(a) 2 5
Yes
(b)  0.28
No
(0  P( E )  1)
(c) 1.09
No
(0  P( E )  1)
Example 2 : Why is the following not a probability model?
Color
Probability
Red
0.28
Green
0.56
Yellow
0.37
Condition 1 :
(0  P( E )  1)
Condition 2 :
( P(ei )  1)
Check :
0.28  0.56  0.37  1.21  1
Condition 2 was not met.
Example 3 : Given :
S  { e 1 , e2 , e3 , e4 }
P(e1 )  P(e2 )  0.2 and P(e3 )  0.5
P(e4 )
Find :
Condition 1 :
Condition 2 :
(0  P( E )  1)
( P(ei )  1)
P(e1 )  P(e2 )  P(e3 )  P(e4 )  1
0.2  0.2  0.5  P(e4 )  1
0.9  P(e4 )  1
P(e4 )  1  0.9
P(e4 )  0.1
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective C : Calculating Probabilities
C1. Approximating Probabilities Using the Empirical Approach
(Relative Frequency Approximation of Probability)
The probability of 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
Example 1 : Suppose that you roll a die 100 times and get six 80
times. Based on these results, what is the probability
that the next roll results in six?
P (six ) 
4
80
 or 0.8
100 5
Example 2 : During a sale at men’s store, 16 white sweaters, 3 red
sweaters, 9 blue sweaters and 7 yellow sweaters were
purchased. If a customer is selected at random, find the
probability that he bought a sweater that was not white.
Frequency of not white  3  9  7  19
Number of trials   f  16  3  9  7  35
19
P(not white ) 
or 0.543 (3 decimal places )
35
Example 3 : The age distribution of employees for this college is
shown below:
Age
# of Employees
Under 20
25
20 – 29
48
30 – 39
32
40 – 49
15
50 and over
10
 f 130
If an employee is selected at random, find the probability that he or
she is in the following age groups.
(a) Between 30 and 39 years of age
32 16

or 0.246 (3 decimal places )
130 65
(b) Under 20 or over 49 years of age
7
25  10 35


or 0.269 (3 decimal places )
130
130 26
C2. Classical Approach to Probability
(Equally Likely Outcomes are required)
If an experiment has n equally likely outcomes and if the number of
ways that an event E can occur in m, then the probability of E ,
P (E ) , is
Number of ways that E can occur m
P( E ) 

Number of possible outcomes
n
If S is the sample space of this experiment,
N (E)
P( E ) 
N (S )
where N (E ) is the number of outcomes in event E , and N (S ) is
the number of outcomes in the sample space.
Example 1 : Let the sample space be S  {1, 2, 3, 4, 5, 6 , 7, 8, 9,10} .
Suppose the outcomes are equally likely.
(a) Compute the probability of the event F  {5, 9} .
N (F )  2
N ( S )  10
N (F )
2
1
P( F ) 


or 0.2
N (S )
10
5
(b) Compute the probability of the event E  “an odd number.”
E  {1, 3, 5, 7, 9}
 N (E)  5
N (E)
5
1
P( E ) 


or 0.5
N (S )
10
2
Example 2 : Two dice are tossed. Find the probability that the sum of
two dice is greater than 8.
N ( S )  36
N ( E )  10
N ( E ) 10 5
E  {( 3,6), (4,5), (4,6), (5,4), (5,5), P( E ) 


or  0.278
N ( S ) 36 18
(5,6), (6,3), (6,4), (6,5), (6,6)}
Example 3 : If one card is drawn from a deck, find the probability of
getting (a) a club; (b) a 4 and a club.
(a) a club
N (a club ) 13 1
P(a club ) 

 or 0.25
N (S )
52 4
(b) a 4 and a club
P(a 4 and a club ) 
N (a 4 and a club ) 1

 0.019
52
N (S )
Example 4 : Three equally qualified runners, Mark, Bill, and Alan, run
a 100-meter sprint, and the order of finish is recorded.
(a) Give a sample space S .
(b) What is the probability that Mark will finish last?
B
A
A
B
M
M
A
A
M
M
B
B
M
B
A
(a) S  {MBA, MAB, BMA, BAM , AMB, ABM }
(b) E  {BAM , ABM }
N (E) 2 1
P( E ) 
  or  0.333
N (S ) 6 3
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Event A and B are disjoint (mutually exclusive) if they have no
outcomes in common.
Addition Rule for Disjoint Events
If E and F are disjoint events, then P( E or F )  P( E )  P( F ) .
Example 1: A standard deck of cards contain 52 cards. One card is
randomly selected from the deck. Compute the probability
of randomly selecting a two or three from a deck of cards.
4
4
8
2
P(2 or 3)  P(2)  P(3) 



or  0.154
52 52 52 13
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective B : General Addition Rule
The General Addition Rule for events that are Not Disjoint
For any two events E and F, P( E or F )  P( E )  P( F )  P( E and F )
Example 1 : A standard deck of cards contain 52 cards. One card is
randomly selected from the deck. Compute the probability
of randomly selecting a two or club from a deck of cards.
P(2 or club )  P(2)  P(club )  P(2 and club )
4
13
1
16
4





or  0.308
52 52 52 52 13
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective C : Complement Rule
Complement Rule
If E represents any event and Ec represents the complement of E,
then
P( E C )  1  P( E )
e.g. The chance of raining tomorrow is 70%. What is the probability
that it will not rain tomorrow?
P(not raining )  1  P(raining )
 1  0.7  0.3
Example 1 : A probability experiment is conducted in which the sample
space of the experiment is S  {1, 2, 3, 4, 5, 6 , 7, 8, 9, 10, 11, 12}
Let event E  {2, 3, 5, 6 , 7} , event F  { 5, 6 , 7, 8}, and
event G  { 9, 11}.
(a) List the outcome in E and F. Are E and F mutually exclusive?
E  {2, 3, 5, 6 , 7}
F  { 5, 6 , 7, 8}
Since there are common elements of 5, 6, and 7 between E and F,
E and F are not mutually exclusive.
(b) Are F and G mutually exclusive? Explain.
Yes, there is no common element between F and G.
(c) List the outcome in E or F. Find P ( E or F ) by counting the
number of outcomes in E or F.
E or F  {2, 3, 5, 6 , 7, 8}
N ( E or F )  6
N ( S )  12
P ( E or F ) 
6
1

or 0.5
12
2
(d) Determine P ( E or F ) using the General Addition Rule.
P( E or F )  P( E )  P( F )  P( E and F )
5
3
6
1
4
 


 or 0.5
12 12 12 12 2
(e) List the outcome in EC. Find P( E C ) by counting the number of
outcomes in EC.
E C  {1, 4, 8, 9 ,10, 11, 12}
C
7
N
(
E
)
C
P( E ) 

12
N (S )
(f) Determine P( E C ) using Complement Rule.
P ( E C )  1  P( E )
7
5 12 5
 1 
 
12 12 12 12
Example 2: In a large department store, there are 2 managers, 4
department heads, 16 clerks, and 4 stock persons. If a
person is selected at random,
(a) find the probability that the person is a clerk or a
manager;
(b) find the probability that the person is not a clerk.
(a) P(clerk or manager )  P(clerk )  P(manager )  P(clerk and manager )

(b)
16
2
0
18
9




 0.692
26 26 26 26 13
P(not a clerk )  1  P(clerk )
16 26 16 10 5
 1



  0.385
26 26 26 26 13
Example 3: The following probability show the distribution for the
number of rooms in U.S. housing units.
Rooms
Probability
One
0.005
Two
0.011
Three
0.088
Four
0.183
Five
0.230
Six
0.204
Seven
0.123
Eight or more
0.156
Source: U.S. Censor Bureau
(a) Verify that this is a probability model.
Is each probability outcome between 0 and 1? Yes
Is ( P(ei )  1) ?
0.005  0.011  0.088  0.183  0.230  0.204  0.123  0.156  1 Yes
(b) What is the probability that a randomly selected housing unit has
four or more rooms? Interpret this probability.
P(4 or more rooms )  0.183  0.230  0.204  0.123  0.156
 0.896
Approximate 89.6% of housing unit has four or more rooms.
Example 4: According the U.S. Censor Bureau, the probability that a
randomly selected household speaks only English at home
is 0.81. The probability that a randomly selected
household speaks only Spanish at home is 0.12.
(a) What is the probability that a randomly selected household speaks
only English or only Spanish at home?
P(ENG only or SPN only )  P(ENG only )  P(SPN only )
 P(ENG only and SPN only )
 0.81  0.12  0  0.93
(b) What is the probability that a randomly selected household speaks
a language other than only English at home?
P(not ENG only )  1  P(ENG only )
 1  0.81  0.19
(c) Can the probability that a randomly selected household speaks only
Polish at home equal to 0.08? Why or why not?
No, because  P(ei )  1
i.e. 0.81  0.12  0.08  1.01  1
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective D : Contingency Table
A Contingency table relates two categories of data. It is also called
a two-way table which consists of a row variable and a column
variable. Each box inside a table is called a cell.
Example 1 : In a certain geographic region, newspapers are classified
as being published daily morning, daily evening and
weekly. Some have a comics section and other do not. The
distribution is shown here.
Have Comics Section
(CY)
Yes
(CN)
No
(M)
Morning
(E)
Evening
(W)
Weekly
2
3
1
6
3
4
2
9
5
7
3
15
If a newspaper is selected at random, find these probabilities.
(a) The newspaper is a weekly publication.
1
N (W ) 3
P(W ) 


N ( S ) 15 5
(b) The newspaper is a daily morning publication or has comics .
P( M or CY )  P( M )  P(CY )  P( M and CY )
5
6
2
9 3
 



15 15 15 15 5
(c) The newspaper is a weekly or does not have comics .
P(W or CN )  P(W )  P(CN )  P(W and CN )
3 9
2 10 2
  


15 15 15 15 3
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Two events are independent if the occurrence of event E does not
affect the probability of event F.
Two events are dependent if the occurrence of event E affects the
probability of event F.
Example 1 : Determine whether the events E and F are independent
or dependent. Justify your answer.
(a) E: The battery in your cell phone is dead.
F: The battery in your calculator is dead.
Independent
(b) E: You are late to class.
F: Your car runs out of gas.
Dependent
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective B : Multiplication Rule for Independent Events
If E and F are independent events, then P( E and F )  P( E )  P( F ) .
Example 1 : If 36% of college students are underweight, find the
probability that if three college students are selected
at random, all will be underweight.
Independent case
P(1st underweight and 2nd underweight and 3rd underweight)
 P(1st underweigh t )  P(2nd underweigh t )  P(3rd underweigh t )
 0.36  0.36  0.36  0.047
Example 2 : If 25% of U.S. federal prison inmates are not U.S. citizens,
find the probability that two randomly selected federal
prison inmates will be U.S. citizens.
Independent case
P(1st U.S. citizen and 2nd U.S. citizen )
 P(1st U.S. citizen )  P(2nd U.S. citizen )
 0.75  0.75  0.563
Section 5.1 Probability Rule
Objective A : Sample Spaces and Events
Objective B : Requirements for Probabilities
Objective C : Calculating Probabilities
Section 5.2 The Additional Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive)
Events
Objective B : General Addition Rule
Objective C : Complement Rule
Objective D : Contingency Table
Section 5.3 Independence and Multiplication Rule
Objective A : Independent Events
Objective B : Multiplication Rule for Independent Events
Objective C : At-Least Probabilities
Objective C : At-Least Probabilities
Probabilities involving the phrase “at least” typically use the
Complement Rule. The phrase at least means “greater than or equal
to.” For example, a person must be at least 17 years old to see an Rrated movie.
Example 1 : If you make random guesses for four multiple-choice test
questions (each with five possible answers), what is the
probability of getting at least one correct?
Direct method :
P(at least one correct )
 P(one correct )  P( two correct )  P( three correct )  P(four correct )
Indirect method :
What is the opposite of at least one correct? None is correct.
4 4 4 4
P (none correct )      
5 5 5 5
4
 
5
4
P(at least one correct )  1  P(none correct )
4
4
 1     0.590
5
Example 2 : For the fiscal year of 2007, the IRS audited 1.77% of
individual tax returns with income of $100,000 or more.
Suppose this percentage stays the same for the current
fiscal year.
(a) Would it be unusual for a return with income of $100,000
or more to be audited?
Yes, 1.77% is unusually low chance of being audited.
(In general, probability of less than 5% is considered to
be unusual.)
(b) What is the probability that two randomly selected returns
with income of $100,000 or more to be audited?
Let A be the return with income of $100,000 or more being
audited.
P(1st A and 2nd A)  (0.0177)  (0.0177)
 0.000313  0
(c) What is the probability that two randomly selected returns with
income of $100,000 or more will NOT be audited?
P(1st not A and 2nd not A)
 (0.9823)  (0.9823)  0.965
(d) What is the probability that at least one of the two randomly
selected returns with income of $100,000 or more to be audited?
P(at least one A)  1  P(none A)
 1  0.965  0.035
Section 5.4 Conditional Probability and the
General Multiplication Rule
Objective A : Conditional Probability and the General
Multiplication Rule
Objective B : Application
Section 5.5 Counting Techniques
Objective A : Multiplication Rule of Counting
Objective B : Permutation or Combination
Objective C : Using the Counting Techniques to Find
Probabilities
Section 5.4 Conditional Probability and the
General Multiplication Rule
Objective A : Conditional Probability and the General
Multiplication Rule
A1. Multiplication Rule for Dependent Events
If E and F are dependent events, then P( E and F )  P( E )  P( F | E ) .
The probability of E and F is the probability of event E occurring
times the probability of event F occurring, given the occurrence of
event E.
Example 1: A box has 5 red balls and 2 white balls. If two balls are
randomly selected (one after the other), what is the
probability that they both are red?
(a) With replacement → Independent case
P(1st red and 2nd red )
5 5 25
 P(1st red )  P(2nd red )   
7 7 49
(b) Without replacement → Dependent case
P(1st red and 2nd red )
 P(1st red)  P(2nd red |1st red )
5 4 20 10
  

7 6 42 21
Example 2: Three cards are drawn from a deck without replacement.
Find the probability that all are jacks.
Without replacement → Dependent case
P(1st jack and 2nd jack and 3rd jack )
4 3 2
1 1 1
1
 

 


 0.00018
52 51 50 13 17 25 5525
(Almost zero percent of a chance)
A2. Conditional Probability
P( E and F ) N ( E and F )
If E and F are any two events, then P( F | E ) 
.

P( E )
N (E)
The probability of event F occurring, given the occurrence of event E,
is found by dividing the probability of E and F by the probability of E.
Example 1 : At a local Country Club, 65% of the members play bridge
and swim, and 72% play bridge. If a member is selected
at random, find the probability that the member swims,
given that the member plays bridge.
Let B be the event of playing bridge.
Let S be the event of swimming.
Given : P( B and S )  0.65
P( B)  0.72
Find : P ( S | B )
0.65
P( S and B)
P( S | B) 

 0.903
P( B)
0.72
Section 5.4 Conditional Probability and the
General Multiplication Rule
Objective A : Conditional Probability and the General
Multiplication Rule
Objective B : Application
Section 5.5 Counting Techniques
Objective A : Multiplication Rule of Counting
Objective B : Permutation or Combination
Objective C : Using the Counting Techniques to Find
Probabilities
Example 1 : Eighty students in a school cafeteria were asked if they
favored a ban on smoking in the cafeteria. The results of
the survey are shown in the table.
(Fa)
(O)
(N)
Class
Favor
Oppose
No opinion
50
(F) Freshman
15
27
8
(S) Sophomore
23
5
2
30
38
32
10
80
If a student is selected at random, find these probabilities.
(a) The student is a freshman or favors the ban.
P( F or Fa)  P( F )  P( Fa)  P( F and Fa)
50 38 15



80 80 80
73

80
(b) Given that the student favors the ban, the student is a sophomore.
P( S and Fa)
P( S | Fa) 
P( Fa)
23 80

38 80
23

38
Example 2 : The local golf store sells an “onion bag” that contains
35 “experienced” golf balls. Suppose that the bag
contains 20 Titleists, 8 Maxflis and 7 Top-Flites.
(a) What is the probability that two randomly selected golf balls are
both Titleists?
Without replacement → Dependent case
P(1st Titleist and 2nd Titleist)
20 19
380
 

 0.319
35 34 1190
(b) What is the probability that the first ball selected is a Titleist and
the second is a Maxfli?
Without replacement → Dependent case
P(1st Titleist and 2nd Maxfli)
20 8
160
 

 0.134
35 34 1190
(c) What is the probability that the first ball selected is a Maxfli and
the second is a Titleist?
Without replacement → Dependent case
P(1st Maxfli and 2nd Titleist)
8 20
160
 

 0.134
35 34 1190
(d) What is the probability that one golf ball is a Titleist and the other
is a Maxfli?
Without replacement → Dependent case
P(1st Titleist and 2nd Maxfli or1st Maxfli and 2nd Titleist)
8 20
20 8
320


 

 0.269
35 34 35 34 1190
Section 5.5 Counting Techniques
Objective B : Permutation and Combination
Permutation
The number of ways we can arrange n distinct objects, taking them
r at one time, is
n!
n Pr 
(n  r)!
Order matters
Combination
The number of distinct combinations of n distinct objects that can
be formed, taking them r at one time, is
n!
n Cr 
r!(n  r)!
Order doesn’t matters
Example 1: Given the letters A, B, C, and D, list the permutations and
combinations for selecting two letters.
The permutations are
AB
BA
CA
DA
AC
BC
CB
DB
AD
BD
CD
DC
Note: AB = BA in combination. Therefore, if duplicates are removed from
a list of permuations, what is left is a list of combinations.
The combinations are
/
AB
BA
AC
BC
AD
BD
/
CB
/
CA
CD
/
DB
/
DC
/
DA
Example 1 : Find (a) 5!
(a) 5!
(b) 25 P8
(c) 12 P4
(c)
 5  4  3  2 1
12
P4
 11880
 120
(b)
P
25 8
 4360910400 0
(d)
25
C8
 1081575
(d) 25 C8
Example 2 : An inspector must select 3 tests to perform in a certain
order on a manufactured part. He has a choice of 7 tests.
How many ways can be performed 3 different tests?
n7
r 3
P  210
7 3
Example 3 : If a person can select 3 presents from 10 presents, how
many different combinations are there?
n  10
10
C3  120
r 3
Section 5.4 Conditional Probability and the
General Multiplication Rule
Objective A : Conditional Probability and the General
Multiplication Rule
Objective B : Application
Section 5.5 Counting Techniques
Objective A : Multiplication Rule of Counting
Objective B : Permutation or Combination
Objective C : Using the Counting Techniques to Find
Probabilities
Section 5.5 Counting Techniques
Objective C : Using the Counting Techniques to Find
Probabilities
After using the multiplication rule, combination and permutation
learned from this section to count the number of outcomes for a
sample space, N (S ) and the number of outcomes for an event,
N (E ) , we can calculate P (E ) by the formula P( E ) 
N (E)
.
N (S )
Example 1 : A Social Security number is used to identify each resident
of the United States uniquely. The number is of the form
xxx-xx-xxxx, where each x is a digit from 0 to 9.
(a) How many Social Security numbers can be formed?
N (S )  10 9  1,000,000,000  one billion
(b) What is the probability of correctly guessing the Social Security
number of the President of the United States?
1
N (E)
1
 one out of a billion
P( E ) 
 9 
N ( S ) 10 1,000,000,000
Example 2 : Suppose that there are 55 Democrats and 45 Republicans
in the U.S. Senate. A committee of seven senators is to be
formed by selecting members of the Senate randomly.
(a) What is the probability that the committee is composed of all
Democrats?
N (S )  100 C7  16, 007,560,800
N ( D) 
55
C7  202,927, 725
202,927, 725
N ( D)
 0.0127
P( D) 

N ( S ) 16, 007,560,800
(b) What is the probability that the committee is composed of all
Republicans?
N (S )  100 C7  16, 007,560,800
C7  45,379, 620
45,379, 620
N ( R)
 0.0028
P( D) 

N ( S ) 16, 007,560,800
N ( R) 
45
(c) What is the probability that the committee is composed of all
three Democrats and four Republicans?
N (S )  100 C7  16, 007,560,800
N (3D and 4 R)  55 C3  45 C4  3,908,883,825
N (3D and 4 R)
3,908,883,825
P( D) 

 0.2442
N (S )
16, 007,560,800
Example 3 :
Five cards are selected from a 52-card deck for a poker hand. (A poker hand
consists of 5 cards dealt in any order.)
(a) How many outcomes are in the sample space?
(b) A royal flush is a hand that contains that A, K, Q, J, 10, all in the same suit.
How many ways are there to get a royal flush?
A, K, Q, J, 10 - ♠
A, K, Q, J, 10 -♥
A, K, Q, J, 10 - ♣
A, K, Q, J, 10 - ♦
n(A) = 4 combinations
(c) What is the probability of being dealt a royal flush?
5.6
Putting It Together:
Which Method Do I Use?
Objectives
1. Determine the appropriate probability rule
to use
2. Determine the appropriate counting
technique to use
Objective 1
• Determine the Appropriate Probability Rule to
Use
5-76