Question

If \( P(A|B) = P(A'|B) \), then which of the following statements is true?

• A. \( P(A) = P(A') \)
• B. \( P(A) = 2P(B) \)
• C. \( P(A \cap B) = \frac{1}{2}P(B) \)
• D. \( P(A \cap B) = 2P(B) \)

Hint:

When \( P(A|B) = P(A'|B) \), use the complement rule and the fact that probabilities sum to 1 to simplify.

Answer 1

The Correct Option is C

Given \( P(A|B) = P(A'|B) \), we use the definitions of conditional probability:\[P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(A'|B) = \frac{P(A' \cap B)}{P(B)}.\]Since \( A \) and \( A' \) are complementary events, we know that the probability of the intersection of \( A \) with \( B \) plus the probability of the intersection of \( A' \) with \( B \) equals the probability of \( B \):\[P(A \cap B) + P(A' \cap B) = P(B).\]From the initial equality \( P(A|B) = P(A'|B) \), we can set the conditional probability expressions equal:\[\frac{P(A \cap B)}{P(B)} = \frac{P(A' \cap B)}{P(B)}.\]Multiplying both sides by \( P(B) \) (assuming \( P(B) eq 0 \)), we get:\[P(A \cap B) = P(A' \cap B).\]Substituting this result back into the equation \( P(A \cap B) + P(A' \cap B) = P(B) \):\[P(A \cap B) + P(A \cap B) = P(B).\]This simplifies to:\[2P(A \cap B) = P(B).\]Finally, dividing by 2 yields:\[P(A \cap B) = \frac{1}{2}P(B).\] Final Answer: \[\boxed{P(A \cap B) = \frac{1}{2}P(B)}\]