Question

P(A1) = .20, P(A2) = .40, P(A3) = .40, P(B1 | A1) = .25, P(B1 | A2) = .05, and P(B1 | A3) = .10. Use Bayes’ theorem to determine P(A3 | B1).

Answer 1

Explanation:

P(A|B) = P(B|A)×P(A)/P(B)

so, in our case

P(A3|B1) = P(B1|A3)×P(A3)/P(B1) = 0.1×0.4/P(B1)

I assume that the problem definition means that A1, A2, A3 are independent and not overlapping events (hence their sum is 1). and therefore, B1 can only happen after A1 or after A2 or after A3. there is no other possibility for B1 to happen.

so,

P(B1) = P(B1|A1) + P(B1|A2) + P(B1|A3) =

= 0.25 + 0.05 + 0.1 = 0.4

therefore,

P(A3|B1) = P(B1|A3)×P(A3)/P(B1) = 0.1×0.4/0.4 = 0.1

Answer 2

Answer:The result follows from manipulating the conditional probability. By definition,

By the law of total probability,

So we have