Probability Examples
Jake Blanchard
Spring 2010
Uncertainty Analysis for Engineers
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Waste in Rivers (Example 2.21)

A chemical plant dumps waste in two
rivers (A and B). We can measure the
contamination in these rivers.
◦ =event that A is contaminated
◦ =event that B is contaminated
◦ P()=0.2; P()=0.33; P()=0.1 (both)

What is probability that at least one river
is contaminated?
Uncertainty Analysis for Engineers
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At least one is contaminated…
P()=P()+P()-P()
 P()=0.2+0.33-0.1=0.43


If B is contaminated, what is the
probability that A is also contaminated?
Uncertainty Analysis for Engineers
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If B, also A…
P(|)= P()/P()
 P(|)=0.1/0.33=0.3


What is the probability that exactly one
river is polluted?
Uncertainty Analysis for Engineers
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Exactly one polluted…

P()-P()=0.43-0.1=0.33
Uncertainty Analysis for Engineers
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Power Plants and Brownouts
(Example 2.22)

We have 2 power plants (A and B)
◦ =failure of A
◦ =failure of B
◦ P()=0.05; P()=0.07; P()=0.01 (both)

If one of the two plants fails, what is
probability the other fails as well?
Uncertainty Analysis for Engineers
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If one fails, the other fails as well…
P(|)= P()/P()=0.01/0.07=0.14
 P(|)= P()/P()=0.01/0.05=0.2


What is the probability of a brownout (at
least one fails)?
Uncertainty Analysis for Engineers
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At least one fails…
P()=P()+P()-P()
 P()=0.05+0.07-0.01=0.11


What is probability that a brownout is
caused by the failure of both plants?
Uncertainty Analysis for Engineers
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Brownout from both failing…
Probability of brownout is 0.11
 Probability of both plants failing is 0.01
 Probability of brownout from both failing
is 0.01/0.11=0.09

Uncertainty Analysis for Engineers
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A Useful Formula
P(D)=P(D|AX)*P(AX)+ P(D|AY)* P(AY)+
P(D|BX)* P(BX)+ P(D|BY)* P(BY)
 This works as long as A, B and X, Y cover
all possible outcomes

Uncertainty Analysis for Engineers
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Highway Congestion (Example 2.26)
I1
I3
I2
Uncertainty Analysis for Engineers
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Highway Congestion (Example 2.26)
E1=congestion on interstate 1
 P(E1)=0.1; P(E2)=0.2
 P(E1|E2)=0.4; P(E2|E1)=0.8 (Bayes’
Theorem)
 P(E3|E1E2)=0.2; ie if no congestion on 1
or 2, then probability of congesion on 3 is
20%
 Also, there is 100% probability of
congestion on 1 if there is congestion on
either 1 or 2

Uncertainty Analysis for Engineers
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Example 2.26
There are 4 possibilities: E1E2, E1E2, E1E2,
E1E2
 P(E1E2)=P(E1|E2)P(E2)=.4*.2=.08
 P(E1E2)=P(E2|E1)P(E1)=(1-.8) *.2=.02
 P(E1E2)=P(E1|E2)P(E2)=(1-.4)*.2=.12
 P(E1E2)=1-.08-.02-.12=.78
 P(E3)=P(E3|E1E2)P(E1E2)+
P(E3|E1E2)P(E1E2)+ P(E3|E1E2)P(E1E2)+
P(E3|E1E2)P(E1E2)=0.376

Uncertainty Analysis for Engineers
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