Q.If $A$ and $B$ are two events and $A \neq \varphi$, $B \neq \varphi$, then
(A) $P(A \mid B) = P(A) \cdot P(B)$
(B) $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
(C) $P(A \mid B) \cdot P(B \mid A) = 1$
(D) $P(A \mid B) = \dfrac{P(A)}{P(B)}$

Concept: Conditional Probability — the probability of one event given that another has occurred.

Step 1: By definition, the conditional probability of $A$ given $B$ is

Step 2: Check each option:

• (A) is the product rule for independent events, not the definition.
• (B) matches the definition exactly.
• (C) is false in general; $P(A \mid B) \cdot P(B \mid A) = \frac{P(A \cap B)^2}{P(A)P(B)}$, which equals 1 only in special cases. …

The core idea here is conditional probability — the probability that event $A$ happens, given that we already know event $B$ has occurred. When we condition on $B$, the sample space effectively shrinks from the entire universe of outcomes to just those outcomes where $B$ occurs. So the probability of $A$ under this new "restricted" space is the proportion of $B$-outcomes that also belong to $A$.

That proportion is simply the fraction $\frac{P(A \cap B)}{P(B)}$, provided $P(B) > 0$. This is not a guess or a convention — it follows directly from the definition of probability in a reduced sample space.

Let’s examine each option one by one.

1. Option (A): $P(A \mid B) = P(A) \cdot P(B)$

   This would mean the conditional probability equals the product of the two individual probabilities. That is almost never true. In fact, $P(A \mid B)$ is a number between 0 and 1, while $P(A) \cdot P(B)$ is typically much smaller. For example, if $P(A) = 0.5$ and $P(B) = 0.5$, the right side is $0.25$, but $P(A \mid B)$ could be anything from 0 to 1. So this is clearly wrong.

2. Option (B): $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$

   This is the definition of conditional probability. It says: out of all outcomes where $B$ happens, what fraction also have $A$ happen? That fraction is exactly the ratio of the overlap to the total of $B$. This is always correct (as long as $P(B) \neq 0$).

3. Option (C): $P(A \mid B) \cdot P(B \mid A) = 1$

   Let’s test this. Using the definition:

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

   Their product is $\frac{P(A \cap B)^2}{P(A) P(B)}$. …

Method: Testing candidate identities against the definition

For "which statement is always true" questions, do not compute a number — instead check each option against the definition and eliminate.

Steps

Step 1: Anchor on the definition.

Any option matching this exactly is the answer.

Step 2: Eliminate the classic look-alikes.

• $P(A)\,P(B)$ is the independence product for $P(A\cap B)$, not for $P(A\mid B)$.
• $\dfrac{P(A)}{P(B)}$ ignores the overlap, so it fails (e.g. disjoint events give $P(A\mid B)=0$). …

Common Mistakes

Mistake 1: Picking $P(A\mid B)=P(A)\,P(B)$.

Why it's wrong: that is the intersection rule for independent events, not the definition of a conditional probability. Correct approach: the definition is always $\dfrac{P(A\cap B)}{P(B)}$.

Mistake 2: Believing $P(A\mid B)\cdot P(B\mid A)=1$. …