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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.

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    $\begingroup$ Somebody must say this: look at the mathematical definitions of independence and conditional probability and stick to them, at present you seem to rely on vague intuitions and this drives you into the wall. Back to solid ground! $\endgroup$ – Did Sep 3 '16 at 8:35
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If $A$ and $B$ are mutually exclusive, then $$ \mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=0 $$ since $A\cap B=\emptyset$. Therefore your answer to the first part is incorrect. Your answer for the second part looks ok.

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Thus, the occurrence of event B does not allot the occurrence of event A and P(A│B) becomes equal to P(B).

No.   That conclusion makes no sense.

If $A$ cannot occur when $B$ occurs, then the probability for $A$ occurring when $B$ occurs is equal to $0$. $$\mathsf P(A\mid B)=0$$

Further the probability for $A$ not occurring when $B$ does not occur is:

$$\mathsf P(A'\mid B') = 1- \dfrac{\mathsf P(A)}{1-\mathsf P(B)}$$

So, if $A,B$ are exclusive, then $\mathsf P(A\mid B)+\mathsf P(A'\mid B')$ can only equal $1$ when $\mathsf P(A)=0$ .


For the Independence case indeed $\mathsf P(A\mid B)+\mathsf P(A'\mid B') = \mathsf P(A)+\mathsf P(A')=1$

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