Q.Compute $P(A|B)$, if $P(B) = 0.5$ and $P(A \cap B) = 0.32$.

Concept: Conditional Probability — the probability of event $A$ given that event $B$ has occurred.

Step 1: Recall the definition of conditional probability:

Step 2: Substitute the given values:

Step 3: Simplify:

Why conditional probability works this way

When we say $P(A|B)$, we are asking: If we already know $B$ happened, what fraction of that $B$-world also contains $A$? The key insight is that conditioning on $B$ shrinks the universe from the whole sample space to just the outcomes where $B$ occurs. So the probability of $A$ in this restricted world is the proportion of $B$ that overlaps with $A$ — which is exactly $\frac{P(A \cap B)}{P(B)}$.

This is not a definition pulled from thin air. It follows from the idea that probabilities must still sum to 1 in the new, smaller universe. Since $P(B)$ is the total “weight” of that universe, we divide the overlap weight $P(A \cap B)$ by it to renormalise.

Step-by-step

1. Identify what is given.

   We know $P(B) = 0.5$ and $P(A \cap B) = 0.32$. The problem asks for $P(A|B)$.

2. Apply the conditional probability formula directly.

   There is no need to find $P(A)$ or any other quantity — the formula only needs the intersection and the conditioning event’s probability.

3. Perform the division.

   $0.32 \div 0.5 = 0.64$. You can think of it as $32/50 = 16/25 = 0.64$.

4. Interpret the result.

   If $B$ occurs, there is a 64% chance that $A$ also occurs. This makes sense because the overlap ($0.32$) is more than half of $B$’s probability ($0.5$).

Method: Applying the conditional-probability definition

Use this whenever you are asked for the probability of one event given another, and you already know (or can find) the joint probability and the probability of the conditioning event.

Steps

Step 1: Identify the conditioning event

The event written after the bar "$\mid$" is assumed to have happened — it becomes the new, shrunken sample space. Note its probability $P(B)$; it must be non-zero.

Step 2: Write the definition

The numerator is the overlap of the two events; dividing by $P(B)$ rescales so the conditioning event carries all the probability.

Step 3: Substitute and sense-check

Insert the known joint probability $P(A\cap B)$ and $P(B)$, then divide. Check the size: because you divide by $P(B)\le 1$, the result satisfies $P(A\mid B)\ge P(A\cap B)$, and it must still stay within $[0,1]$.

Common Mistakes

Mistake 1: Reporting $P(A\mid B)=P(A\cap B)=0.32$ without dividing by $P(B)$.

Why it's wrong: conditioning rescales to the world where $B$ is certain, so you must divide by $P(B)$. Correct approach: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.32}{0.5}=0.64$.

Mistake 2: Dividing the wrong way, $\dfrac{P(B)}{P(A\cap B)}=\dfrac{0.5}{0.32}$.

Why it's wrong: that gives $1.5625>1$, impossible for a probability — a quick sanity flag. The intersection is always the numerator. Correct approach: $\dfrac{0.32}{0.5}=0.64$.