Download Conditional Probability and Bayes` Theorem

BOBBY B. LYLE
SCHOOL OF ENGINEERING
EMIS - SYSTEMS ENGINEERING PROGRAM
SMU
EMIS 7370 STAT 5340
Department of Engineering Management, Information and Systems
Probability and Statistics for Scientists and Engineers
Conditional Probability and
Bayes’ Theorem
Dr. Jerrell T. Stracener,
SAE Fellow
Leadership in Engineering
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Conditional Probability
Basic Concept
• Definition
• Reduced Sample Space
• Rules
• Bayes’ Rule
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Conditional Probability
If A and B are any events in S and P(B)  0, the
conditional probability of A, given that B has occurred
is denoted by P(A | B), and
P(A  B)
P(A | B) 
P(B)
Note: The given event is called the reduced sample space.
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Conditional Probability: Rules
• Rule
If A and B are any events in S, then
P(A  B) = P(A)P(B|A)
if P(A)  0
= P(B)P(A|B)
if P(B)  0
• Rule
Two events A and B are independent if
P(A|B) = P(A),
and are dependent otherwise.
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Product Rule continued
Rule
If A, B and C are events in S for which P(A) > 0, P(B) > 0,
and P(C) > 0 , then
P(A  B  C) = P(A)P(B|A)P(C|A  B)
•Rule
For events A1, A2, ... An in S, can occur, then
P(A1  A2  ...  An) = P(A1)P(A2|A1)P(A3|A1  A2)…
P(An|A1 ... An-1)
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Bayes Theorem
Let {B1, B2, ..., Bn} be a set of events forming a partition of the
sample space S, where P(Bi)  0, for i = 1, 2, ... , n. Let A be any
event of S such that P(A)  0. Then, for k = 1, 2, ..., n,
P(B k | A) 
P(Bk  A)
n
 P(B  A)
i 1

i
P(B k )P(A | B k )
n
 P(B )P(A | B )
i 1
i
i
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In a sense, Bayes’ Rule is updating or revising the prior
probability P(B) by incorporating the observed information
contained within event A into the model.
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Example
A chain of video stores sells three different brands of VCR’s. Of
its VCR sales, 50% are Brand 1 (the least expensive), 30% are
Brand 2, and 20% are Brand 3. Each manufacturer offers a 1-year
warranty on parts and labor. It is known that 25% of Brand 1’s
VCR’s require warranty repair work, whereas the corresponding
percentages for Brands 2 and 3 are 20% and 10% respectively.
1. What is the probability that a randomly selected purchaser has
bought a Brand 1 VCR that will need repair while under warranty?
2. What is the probability that a randomly selected purchaser has
a VCR that will need repair while under warranty?
3. If a customer returns to the store with a VCR that needs warranty
repair work, what is the probability that it is a Brand 1 VCR? A
Brand 2 VCR? A Brand 3 VCR?
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Example: solution
The probability that a VCR sold will be Brand 1 is P(B1) = 0.50,
the probability it will be Brand 2 is P(B2) = 0.30, and the probability
that it will be Brand 3 P(B3) = 0.20.
Once a Brand of VCR has been selected
R represents that the VCR needs repair
R’ represents that the VCR does not need repair
The probability that a Brand 1 VCR needs repair, P(R|B1) = 0.25
The probability that a Brand 2 VCR needs repair, P(R|B2) = 0.20
The probability that a Brand 3 VCR needs repair, P(R|B3) = 0.10
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Example: solution
Outcome
0.25
B1R
0.125
no repair 0.75
B1R'
0.375
repair
B2R
0.060
B2R'
0.240
0.10
B3R
0.020
0.90
B3R'
0.180
repair
Brand 1
0.50
Probability
Brand 2
0.30
0.20
no repair 0.80
0.20
Brand 3
repair
no repair
1.000  10
Example: solution
1. P(B1 and R) = P(B1  R)
= P(B1)P(R|B1)
= (0.50)(0.25)
= 0.125
or, by inspection from the tree diagram P(B1 and R) = 0.125
2. Since R = (B1  R)  (B2  R)  (B3  R),
P(R) = P(B1  R) + P(B2  R) + P(B3  R)
= 0.125 + 0.060 + 0.020
= 0.205
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Example: solution
3.
P(B1|R) = P(B1  R)/P(R) = 0.125/0.205 = 0.61
P(B2|R) = P(B2  R)/P(R) = 0.060/0.205 = 0.29
P(B3|R) = P(B2  R)/P(R) = 0.020/0.205 = 0.1
Note:
P(B3|R) = 1 - P(B1|R) - P(B2|R) = 0. 10
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Example
An electrical system consists of four components whose
reliability configuration is
C
A
B
D
The system works if components A and B work and either of the
components C or D work. The reliability (probability of working)
of each component is 0.9.
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Find the probability that:
(a) The entire system works
(b) The component C does not work, given that the entire system
works.
Assume that four components work independently.
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Solution
In this configuration of the system, A, B, and the subsystem C and
D constitute a serial circuit system, whereas the subsystem C
and D itself is a parallel circuit system.
(a) Clearly the probability that the entire system works can be
calculated as the following:
P( A  B  (C  D))  P( A) P( B) P(C  D)
 P( A) P( B)[1  P(C ' D' )]
 P( A) P( B)[1  P(C ' ) P( D' )]
 (0.9)(0.9)[1  (1  0.9)(1  0.9)]
 0.8019
The equalities above hold because of the independence among the
four components
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(b) To calculate the conditional probability in this case, notice that
P(the system works but C does not work )
P=
P(the system works)
(0.9)(0.9)(1-0.9)(0.9)
P ( A  B  C ' D)
= P(the system works)
=
=0.090909
0.8019
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