If A and B are events such that \(P(A|B)=P(B|A)\) and \(P(A \cap B) \ne 0\), then :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(P(A)=P(B)\)
By the definition of conditional probability:
\( \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)},\quad P(B|A) = \frac{P(A \cap B)}{P(A)} \)
Equating the two:
\( \displaystyle \frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)} \)
Since \(P(A \cap B) \ne 0\), it can be cancelled:
\( \displaystyle \frac{1}{P(B)} = \frac{1}{P(A)} \)
\(P(A) = P(B)\)
Why the other options are incorrect
Option (1) \(A=B\) — Incorrect. The events need not be equal. Throw a die with A = {1, 2} and B = {2, 3}: \(P(A|B) = \frac{1}{2} = P(B|A)\) and \(P(A \cap B) = \frac{1}{6}\), but A ≠ B.
Option (2) \(A \cap B=\phi\) — Incorrect. Then \(P(A \cap B)\) would be 0, which the question rules out.
Option (3) \(A \subset B\) but \(A \ne B\) — Incorrect. With A = {1, 2} and B = {2, 3} on a die, \(P(A|B) = \frac{1}{2} = P(B|A)\), yet A is not a subset of B.
Note: NTA dropped this question. For the benefit of students and to give them additional practice, the question has been suitably modified so that it has one clear answer.