**Question:**

Let A and B be events where 0 < P(A) < 1 and 0 < P(B) < 1. Is P(A|B) + P(A' |B' ) = 1 true when A and B are • mutually exclusive? • independent? If the answer is negative, provide a counterexample.

**My Answer:**

*Mutually Exclusive*

When two events are mutually exclusive, the occurrence of the first event does not allow the second event to occur. If my car totally breaks down, there is no possibility for me to be at class on time (assuming I only drive to school by my own car). Thus, the occurrence of event B does not allot the occurrence of event A and P(A│B) becomes equal to P(B). Similarly, P(A'│B') becomes equal to P(B'). Finally, the formulation given in the question becomes as following: P(B)+P(B') = 1, which is true because the total probability of an events occurring or not occurring is 1.

*Independent*

According to the definition: If P(B) ≠ 0, then A and B are independent if only if P(A│B) = P(A), Assuming that A and B are independent by seeing 0 < P(B) < 1 (which denotes P(B) is absolutely larger than 0), then P(A│B) = P(A). Besides, assuming A and B are independent, we can say P(A'│B') = P(A'). So we can rewrite the formulation given in the question as: P(A) + P(A') = 1 Shortly, when A and B are independent this means B or B' doesn’t affect the result of A or Ac. Thus, the formulation given in the question is true when A and B are independent is true.

*Is my answer correct?*

Thanks for comments.