Q.Let $A$ and $B$ be two events such that $P(A) = \dfrac{3}{8}$, $P(B) = \dfrac{5}{8}$ and $P(A \cup B) = \dfrac{3}{4}$. Then $P(A \mid B) \cdot P(A' \mid B)$ is equal to
(A) $\dfrac{2}{5}$
(B) $\dfrac{3}{8}$
(C) $\dfrac{3}{20}$
(D) $\dfrac{6}{25}$

Concept: Conditional Probability

We are given $P(A) = \frac{3}{8}$, $P(B) = \frac{5}{8}$, and $P(A \cup B) = \frac{3}{4}$.

Step 1: Find $P(A \cap B)$

Using the addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P(A \cap B)$

$\frac{3}{4} = 1 - P(A \cap B)$

So $P(A \cap B) = 1 - \frac{3}{4} = \frac{1}{4}$.

Step 2: Compute $P(A \mid B)$ and $P(A' \mid B)$

$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{5/8} = \frac{1}{4} \cdot \frac{8}{5} = \frac{2}{5}$. …

We start with the given probabilities:

$P(A) = \frac{3}{8}$, $P(B) = \frac{5}{8}$, and $P(A \cup B) = \frac{3}{4}$.

The problem asks for $P(A \mid B) \cdot P(A' \mid B)$. This is a product of two conditional probabilities under the same condition $B$. The intuition: once we know $B$ has occurred, the probability of $A$ and its complement $A'$ must add to 1. So their product is maximized when they are balanced — but we need the exact numbers.

Step 1: Find $P(A \cap B)$ using the addition rule.

The formula for the union of two events is:

Substitute the known values:

So:

Step 2: Compute $P(A \mid B)$.

By definition:

Step 3: Compute $P(A' \mid B)$. …

Method: Complementary conditionals under a fixed condition

Use this when a product or sum mixes $P(A\mid B)$ and $P(A'\mid B)$ under the same condition $B$.

Steps

Step 1: Recover the overlap and the first conditional.

Get $P(A\cap B)$ from the addition rule (or a given conditional), then

Step 2: Use the fixed-condition complement rule.

Given $B$, the outcomes $A$ and $A'$ partition the reduced sample space, so

$P(A'\mid B)=1-P(A\mid B).$ …

Common Mistakes

Mistake 1: Computing $P(A'\mid B)$ from scratch instead of using $1-P(A\mid B)$.

Why it's wrong: not a fatal error, but it wastes time and invites slips; under the same condition $B$, the events $A$ and $A'$ partition the reduced space. Correct approach: $P(A'\mid B)=1-P(A\mid B)$.

Mistake 2: Forgetting to recover $P(A\cap B)$ from the addition rule. …