Q.If A and B are events such that $P(A|B) = P(B|A)$, then (A) $A \subset B$ but $A \neq B$ (B) $A = B$ (C) $A \cap B = \phi$ (D) $P(A) = P(B)$

From the definition, with $P(A)>0$ and $P(B)>0$ (needed for both conditionals to exist):

Set them equal and cross-multiply: …

Start from the definitions

For $P(A\mid B)$ and $P(B\mid A)$ to be defined we need $P(B)>0$ and $P(A)>0$. Then

Both fractions have the same numerator, $P(A\cap B)$.

Use the given equality

We are told $P(A\mid B)=P(B\mid A)$, so

Cross-multiplying,

In the non-degenerate case $P(A\cap B)\neq 0$ (the events actually overlap), divide it out to get

Why the other options fail …

Method: Comparing Two Conditional Probabilities via Their Definitions

Use this when a condition ties $P(A\mid B)$ to $P(B\mid A)$ (or any two conditionals that share a numerator) and you must deduce what that condition forces about the events.

Steps

Step 1: Expand each conditional with the definition

The key structural observation is that both fractions have the same numerator, $P(A\cap B)$.

Step 2: Impose the given relation and cancel …

Common Mistakes

Mistake 1: Concluding $A=B$ instead of $P(A)=P(B)$.

The symmetric-looking condition tempts students to pick option (B). Why it's wrong: equal conditional probabilities constrain only the probabilities, not the events — two genuinely different events can share $P(A)=P(B)$. Correct approach: write both conditionals and cancel the common numerator $P(A\cap B)$, leaving $P(A)=P(B)$.

Mistake 2: Missing that both conditionals share the same numerator. …