Over the life of a building the probability of excessive set

Over the life of a building, the probability of excessive settlement of the foundation is estimated to be 0.002, whereas the probability of collapse of the superstructure, if that settlement doesn’t occur, is 0.0001. However, if there is excessive settlement of the foundation, the probability of superstructure collapse will increase to 0.006. What is the probability that the superstructure will collapse during the life of the building? (Enter your answer to FIVE decimal places, 0.XXXXX)

Please show work, not just the answer. I want to be able to understand the concepts used to solve the problem. Will rate/review. Thanks. :)

Solution

Let

E = excessive settlement
C =collapse

Hence, by Bayes\' Rule,

P(C) = P(E) P(C|E) + P(E\')P(C|E\')

= 0.002*0.006 + (1-0.002)*(0.0001)

= 0.0001118 = 0.00011 [ANSWER]