How to show that P(B|A')+P(B'|A')=1?

Question

Suppose that A and B are any two events and P(A)>0, show that P(B|A')+P(B'|A')=1?

Answer 1

The condition needs to be that P(A') > 0.

P(B|A') = P(BA')/P(A')

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

Sets B'A' and BA' are disjoint, as B'A'BA' = B'BA' is the null set.

Therefore, P(B'A') + P(BA') = P((B U B')A') = P(A').

P(B|A') + P(B'|A') = P(BA')/P(A') + P(B'A')/P(A') = (P(BA')+ P(B'A'))/P(A') = P(A')/P(A') = 1