Probability Examples Jake Blanchard Spring 2010 Uncertainty Analysis for Engineers1

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

At least one is contaminated… P(  )=P(  )+P(  )-P(  ) P(  )= =0.43 If B is contaminated, what is the probability that A is also contaminated? Uncertainty Analysis for Engineers3

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

Exactly one polluted… P(  )-P(  )= =0.33 Uncertainty Analysis for Engineers5

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

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

At least one fails… P(  )=P(  )+P(  )-P(  ) P(  )= =0.11 What is probability that a brownout is caused by the failure of both plants? Uncertainty Analysis for Engineers8

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

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

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

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

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)= =.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