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Probability Examples Jake Blanchard Spring 2010 Uncertainty Analysis for Engineers1

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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 Engineers2

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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 Engineers3

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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 Engineers4

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Exactly one polluted… P( )-P( )=0.43-0.1=0.33 Uncertainty Analysis for Engineers5

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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 Engineers6

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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 Engineers7

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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 Engineers8

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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 Engineers9

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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 Engineers10

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Highway Congestion (Example 2.26) Uncertainty Analysis for Engineers11 I1I1 I3I3 I2I2

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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 Engineers12

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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 Engineers13

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