Q.A and B are two events such that $P(A) \neq 0$. Find $P(B|A)$, if

Concept: Conditional Probability — $P(B|A) = \frac{P(A \cap B)}{P(A)}$.

Step 1: Recall the definition:

Step 2: For (i), $A \subseteq B$ implies $A \cap B = A$.

Thus $P(A \cap B) = P(A)$, so

Step 3: For (ii), $A \cap B = \phi$ implies $P(A \cap B) = 0$.

Thus

The core idea here is conditional probability — the probability that event $B$ occurs, given that we already know event $A$ has occurred. The formula is:

The denominator $P(A)$ is non-zero (given), so the fraction is well-defined. The numerator is the probability that both $A$ and $B$ happen. The key is to figure out what $A \cap B$ looks like in each case.

Let’s go case by case.

Case (i): $A$ is a subset of $B$

If $A \subseteq B$, then every outcome in $A$ is also in $B$. That means the overlap $A \cap B$ is simply $A$ itself — there is no part of $A$ that lies outside $B$.

So:

Plug this into the formula:

Case (ii): $A \cap B = \phi$

Here, $A$ and $B$ are disjoint — they have no outcomes in common. So the intersection is empty:

Substitute:

Method: Evaluating a conditional probability from the set relationship

For questions that give you how two events sit relative to each other (subset, disjoint, overlapping) rather than numbers, work straight from the definition and reduce the intersection using that relation.

Steps

Step 1: Start from the definition.

Everything hinges on identifying $A \cap B$.

Step 2: Replace $A \cap B$ using the given relationship.

Translate the words into what the overlap must be:

• If $A \subseteq B$, every outcome of $A$ lies in $B$, so $A \cap B = A$.
• If $A \cap B = \varnothing$ (disjoint), the overlap is empty, so $P(A \cap B) = 0$.

Step 3: Substitute and simplify.

$A \subseteq B$ gives $\dfrac{P(A)}{P(A)} = 1$; disjoint gives $\dfrac{0}{P(A)} = 0$.

The reasoning, not arithmetic, is the point: $P(B\mid A)$ measures how much of $A$ also lies in $B$ — total overlap gives $1$, no overlap gives $0$.

Common Mistakes

Mistake 1: Thinking $P(B\mid A)$ is undefined when $A$ and $B$ are disjoint.

Why it's wrong: the formula is perfectly defined since $P(A)\neq 0$; the numerator $P(A\cap B)$ is simply $0$. Correct approach: $P(B\mid A)=\dfrac{0}{P(A)}=0$.

Mistake 2: Getting the subset case backwards.

Why it's wrong: when $A\subseteq B$, whenever $A$ occurs $B$ must occur, so the probability is $1$, not $0$. Correct approach: $A\cap B=A$, giving $P(B\mid A)=\dfrac{P(A)}{P(A)}=1$.