Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
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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