Question

P(A) = 0.10, P(B | A)= 0.39, and P(B | A’) = 0.39. Find the following: P(A’), P(B’ | A), P(B’ | A’), P(B), P(B’), P(A | B), P(A’ | B), P(A | B’), P(A’ | B’).

How would you describe the relationship between Outcomes A and B?

Answer 1

P(A') = 1 - P(A) = 1 - 0.10 = 0.90

Using Baye's rule: P(B) = P(B|A) x P(A) + P(B|A') x P(A') = 0.39*0.10 + 0.39*0.90 = 0.39

So, P(B') = 1 - P(B) = 1 - 0.39 = 0.61

P(B|A) = 0.39 = P(A.B)/P(A)

So, P(A.B) = P(B.A) = P(A) x P(B|A) = 0.10 x 0.39 = 0.039

Use the complementary rule of conditional probability:  P(B'|A)= 1- P(B|A) = 1 - 0.39 = 0.61

And, P(B'|A')= 1- P(B|A') ?= 1 - 0.39 = 0.61

P(A|B) = P(A.B)/P(B) = 0.039/0.39 = 0.10

P(B|A') = 0.39 = P(B.A')/P(A')

So, P(B.A') = P(A'.B) = P(B|A') x P(A') = 0.39 x 0.90 = 0.351

P(A'|B) = P(A'.B)/P(B) = 0.351/0.39 = 0.90

P(A|B') = P(A.B')/P(B') = P(B'|A) x P(A) / P(B') = 0.61*0.10/0.61 = 0.10

P(A'|B') = 1 - P(A|B') = 1 - 0.10 = 0.90

Note that P(A) x P(B) = 0.10 x 0.39 = 0.039 and again P(A.B) = 0.039. So, P(A.B) = P(A) x P(B). This happens for independent events and hence A and B are independent events.