Q.If $P(A) = \dfrac{3}{10}$, $P(B) = \dfrac{2}{5}$ and $P(A \cup B) = \dfrac{3}{5}$, then $P(B \mid A) + P(A \mid B)$ equals
(A) $\dfrac{1}{4}$
(B) $\dfrac{1}{3}$
(C) $\dfrac{5}{12}$
(D) $\dfrac{7}{12}$

Find $P(A\cap B)$ from the addition rule, then the two conditional probabilities.

$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\tfrac{3}{10}+\tfrac{4}{10}-\tfrac{6}{10}=\tfrac{1}{10}.$

$P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{1/10}{3/10}=\tfrac13,$ and $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{1/10}{4/10}=\tfrac14.$ …

What we need

Both conditional probabilities depend on the overlap $P(A\cap B)$, which is not given directly. We recover it from the addition rule.

Step 1 — find $P(A\cap B)$

Working in tenths ($P(A)=\tfrac{3}{10}$, $P(B)=\tfrac25=\tfrac{4}{10}$, $P(A\cup B)=\tfrac35=\tfrac{6}{10}$):

Step 2 — the two conditional probabilities

Step 3 — add …

Method: Recover the intersection first, then form the conditionals

Use this when a conditional probability (or a sum of them) is asked but the overlap $P(A\cap B)$ is not given directly — instead you are told $P(A)$, $P(B)$ and $P(A\cup B)$.

Steps

Step 1: Extract the intersection from the addition rule.

Step 2: Form each conditional probability from the definition.

$P(B\mid A)=\frac{P(A\cap B)}{P(A)},\qquad P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$ …

Common Mistakes

Mistake 1: Trying to compute the conditionals without first finding $P(A\cap B)$.

Why it's wrong: the overlap isn't given directly; skipping it leaves you guessing. Correct approach: recover it from $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ before anything else.

Mistake 2: Assuming $P(B\mid A)=P(A\mid B)$. …