Q.If $P(A|B) > P(A)$, then which of the following is correct : (A) $P(B|A) < P(B)$ (B) $P(A \cap B) < P(A) \cdot P(B)$ (C) $P(B|A) > P(B)$ (D) $P(B|A) = P(B)$

Concept: Conditional Probability — the condition $P(A|B) > P(A)$ means that knowing $B$ has occurred makes $A$ more likely. This implies a positive association between $A$ and $B$.

Step 1: Write the definition of conditional probability:

Step 2: Multiply both sides by $P(B)$:

This eliminates option (B) immediately. …

Why this works: the logic of conditional probability

Conditional probability tells us how the chance of one event changes when we know another event has happened. The statement $P(A|B) > P(A)$ says: knowing B makes A more probable. That’s a statement about a positive association between A and B.

The key insight: if A is more likely in the presence of B, then B must also be more likely in the presence of A. This symmetry is built into the definition of conditional probability — it’s not an assumption, it’s a mathematical consequence.

Let’s see why.

Step-by-step reasoning

1. Write the given condition in terms of the intersection.

By definition:

The condition $P(A|B) > P(A)$ becomes:

2. Multiply both sides by $P(B)$ (which is positive).

This is a clean, useful form: the joint probability exceeds the product of the marginals. That’s the mathematical signature of a positive association.

3. Now examine $P(B|A)$.

By definition:

We already know $P(A \cap B) > P(A) \cdot P(B)$. Substitute this into the numerator:

$P(B|A) > \frac{P(A) \cdot P(B)}{P(A)}$ …

Method: Turning a conditional-probability inequality into the joint form

Use this whenever a question states an inequality between a conditional probability and a plain probability (e.g. $P(A\mid B) > P(A)$) and asks what else must be true.

Steps

Step 1: Replace every conditional probability by its definition.

Never reason about $P(A\mid B)$ directly — rewrite it using

This converts the statement into one about the joint probability $P(A\cap B)$.

Step 2: Clear the denominator to reach the symmetric form.

Because a conditional probability is only defined when $P(B) > 0$, you may multiply through by $P(B)$ without flipping the inequality. The given inequality becomes a statement comparing $P(A\cap B)$ with the product $P(A)\,P(B)$. This form is symmetric in $A$ and $B$. …

Common Mistakes

Mistake 1: Assuming $P(A\mid B)$ and $P(B\mid A)$ move in opposite directions.

Why it's wrong: they are linked symmetrically through $P(A\cap B)$, so if $B$ makes $A$ more likely, then $A$ makes $B$ more likely too. Correct approach: reduce both conditionals to the joint form $P(A\cap B)$ vs $P(A)P(B)$ and you will see they agree, giving option (C), not (A).

Mistake 2: Reading $P(A\mid B) > P(A)$ as $P(A\cap B) < P(A)P(B)$. …