If A and B are two distinct events such that P(A|B) = P(B|A), then which of the following is /are possible?

(A) A= B

(B) P (A) = P(B)

(C) A ⊂ B but A ≠ B

(D) A∩ B = ɸ

Choose the correct answer from the options given below:

Given that A and B are two distinct events (A ≠ B) and P(A|B) = P(B|A).

Using the conditional probability formula:

P(A|B) = $\frac{P(A \cap B)}{P(B)}$

P(B|A) = $\frac{P(A \cap B)}{P(A)}$

Since P(A|B) = P(B|A):

$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$

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$P(A \cap B) \times P(A) = P(A \cap B) \times P(B)$

$P(A \cap B) \times [P(A) - P(B)] = 0$

This equation is satisfied when either:

Case 1: $P(A \cap B) = 0$ (the events are disjoint)

Case 2: $P(A) = P(B)$ (the events have equal probability)

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Checking option (A): A = B

The question states A and B are distinct events, so A ≠ B.

Not possible.

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Checking option (B): P(A) = P(B)

This is Case 2 from above.

Possible.

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Checking option (C): A ⊂ B but A ≠ B

If A is a proper subset of B, then $A \cap B = A$, so $P(A \cap B) = P(A)$.

The condition becomes:

$\frac{P(A)}{P(B)} = \frac{P(A)}{P(A)} = 1$

This means $P(A) = P(B)$.

However, if A ⊂ B and A ≠ B (proper subset), then $P(A) < P(B)$, which contradicts $P(A) = P(B)$.

Not possible.

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Checking option (D): $A \cap B = \phi$ (disjoint events)

This is Case 1 from above. When A and B don't overlap:

$P(A \cap B) = 0$

$P(A|B) = \frac{0}{P(B)} = 0$

$P(B|A) = \frac{0}{P(A)} = 0$

So $P(A|B) = P(B|A) = 0$

Possible.

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Only statements (B) and (D) are possible.