Section 5.5 Counting Techniques

Report
Modular 9
Ch 5.4 to 5.5
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.
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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
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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.
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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 redand2ndred)
5 5 25
 P(1st red)  P(2ndred)   
7 7 49
(b) Without replacement → Dependent case
P(1st redand2ndred)
 P(1st red) P(2ndred |1st red)

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5 4 20 10

 
7 6 42 21
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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 and3rd jack)
4 3 2
1 1 1
1
 

 

 0.00018

52 51 50 13 17 25 5525
(Almost zero percent of a chance)
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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.
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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
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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
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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
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(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
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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
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(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)
160
8 20

 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
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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
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Section 5.5 Counting Techniques
Objective A : Multiplication Rule of Counting
If Task 1 can be done in n ways and Task 2 can be done in m ways,
Task 1 and Task 2 performed together can be done together in
n m ways.
Example 1 : Two dice are tossed. How many outcomes are in the
sample space.
6 6
 36
Example 2 : A password consists of two letters followed by one digit.
How many different passwords can be created? (Note:
Repetitions are allowed)
26  26 10
 6760
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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 and Combination
Objective C : Using the Counting Techniques to Find
Probabilities
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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)!
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Order doesn’t matters
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Example 1 : Find (a) 5!
(a) 5!
(b) 25 P8
(c) 12 P4
(c)
 5  4  3  2 1
12
(d) 25 C8
P4
 11880
 120
(b)
P
25 8
 43609104000
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(d)
25
C8
 1081575
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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
7
r 3
P3  210
Example 3 : If a person can select 3 presents from 10 presents, how
many different combinations are there?
n  10
10
r 3
C3  120
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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
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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 ) 
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N (E)
.
N (S )
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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 )  109  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
P( E ) 
 9 
 one out of a billion
N ( S ) 10 1,000,000,000
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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
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(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 4R)  55 C3  45 C4  3,908,883,825
P( D) 
N (3D and 4 R)
3,908,883,825

 0.2442
N (S )
16, 007,560,800
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