Q.If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct? (A) $P(A|B) = \frac{P(B)}{P(A)}$ (B) $P(A|B) < P(A)$ (C) $P(A|B) \geq P(A)$ (D) None of these

Concept: Conditional Probability — when one event is a subset of another, the conditional probability is at least as large as the unconditional probability of the subset.

Step 1: Since $A \subset B$, every outcome in $A$ is also in $B$. Therefore $A \cap B = A$.

Step 2: By definition, $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$. …

Why This Problem Is About "Narrowing the Sample Space"

The core idea here is conditional probability — the chance of $A$ given that $B$ has already happened. When $A \subset B$, every outcome in $A$ is also in $B$. So if you know $B$ occurred, you've effectively shrunk the possible universe to just $B$. Since $A$ is entirely inside that smaller universe, its relative size inside $B$ should be larger than its size in the original full space.

Let's make this precise.

Step-by-Step Reasoning

1. Recall the definition of conditional probability.

For any two events $A$ and $B$ with $P(B) \neq 0$:

This is the fraction of $B$'s probability that also belongs to $A$.

2. Apply the subset condition $A \subset B$.

If $A$ is a subset of $B$, then every outcome in $A$ is automatically in $B$. That means:

So the intersection is just $A$ itself. The formula simplifies to:

3. Compare $P(A|B)$ with $P(A)$.

We now have:

Since $P(B)$ is a probability, $0 < P(B) \leq 1$ (and $P(B) \neq 0$ is given). Therefore:

Multiplying both sides by $P(A)$ (which is non-negative):

That is:

4. When does equality happen? …

Method: Conditional probability when one event is a subset of the other

Use this whenever you are told $A\subset B$ (or can see one event forces the other) and must compare $P(A\mid B)$ with $P(A)$.

Steps

Step 1: Simplify the intersection using the subset relation.

If $A\subset B$, every outcome of $A$ lies in $B$, so

Step 2: Compare against $P(A)$ using the size of $P(B)$. …

Common Mistakes

Mistake 1: Forgetting that $A\subset B$ makes $A\cap B = A$.

Why it's wrong: without this simplification students cannot reduce $P(A\mid B)$ and may pick "None of these." Correct approach: use $A\cap B=A$ so $P(A\mid B)=P(A)/P(B)$.

Mistake 2: Inverting the ratio to $P(B)/P(A)$. …