Given: \(P(A|B) = P(A'|B)\)
Using the definition of conditional probability, we have:
\(\frac{P(A \cap B)}{P(B)} = \frac{P(A' \cap B)}{P(B)}\)
Since \(P(B)\) is in the denominator on both sides, we can multiply both sides by \(P(B)\), assuming \(P(B) \neq 0\):
\(P(A \cap B) = P(A' \cap B)\)
We know that \(A' = \Omega - A\), where \(\Omega\) is the sample space. Therefore, \(A' \cap B = B - (A \cap B)\). So, we can write \(P(A' \cap B) = P(B) - P(A \cap B)\)
Substituting this back into the equation, we get:
\(P(A \cap B) = P(B) - P(A \cap B)\)
Adding \(P(A \cap B)\) to both sides:
\(2 P(A \cap B) = P(B)\)
Dividing both sides by 2:
\(P(A \cap B) = \frac{1}{2} P(B)\)
Correct Answer: \(P(A\cap B)=\frac{1}{2}P(B)\)
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