Probability Question - None

Question

Let $A$, $B$, and $C$ be three events, which are pair-wise independent and $\bar{E}$ denotes the complement of an event $E$. If $P(A \cap B \cap C) = 0$ and $P(C) > 0$, then $P[(\bar{A} \cap \bar{B}) \mid C]$ is equal to

Accepted Answer

Step 1: Use pairwise independence. We have P(A∩B) = P(A)P(B), P(B∩C) = P(B)P(C), and P(A∩C) = P(A)P(C).

Step 2: Since P(A∩B∩C) = 0 and P(C) > 0, we have P(A∩B|C) = P(A∩B∩C)/P(C) = 0. This means within event C, events A and B cannot both occur.

Step 3: From pairwise independence: P(A∩B∩C) = P(A∩B)·P(C|A∩B). But by direct computation using independence: if both A and B occur with C independently, we'd have P(A)P(B)P(C). Setting this to 0 requires either P(A) = 0 or P(B) = 0.

Step 4: Without loss of generality (by symmetry and the constraint), at least one of P(A∩C) or P(B∩C) must equal 0. Combined with pairwise independence, this gives us P(A∩C) = 0 or P(B∩C) = 0.

Step 5: Compute P[(ar{A}∩ar{B})|C] = P(ar{A}∩ar{B}∩C)/P(C) = [P(C) - P((A∪B)∩C)]/P(C) = [P(C) - P(A∩C) - P(B∩C) + P(A∩B∩C)]/P(C).

Step 6: Since P(A∩B∩C) = 0 and from the constraint one of P(A∩C) or P(B∩C) is 0, and using pairwise independence relations, this simplifies to P[(ar{A}∩ar{B})|C] = P(ar{A})P(ar{B}) or 1 - P(A) - P(B) + P(A)P(B) depending on the specific values.

∴ Answer: A

Question:

Step-by-Step Solution

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