Q.If $A$ and $B$ are two events such that $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$, $P(A \mid B) = \dfrac{1}{4}$, then $P(A' \cap B')$ equals
(A) $\dfrac{1}{12}$
(B) $\dfrac{3}{4}$
(C) $\dfrac{1}{4}$
(D) $\dfrac{3}{16}$

Concept: Conditional Probability — we use $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ to find the intersection, then apply De Morgan’s law for the complement.

Step 1: From $P(A \mid B) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$,

$P(A \cap B) = P(A \mid B) \cdot P(B) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$.

Step 2: We need $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.

Step 3: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12}$.

Compute: $\frac{1}{2} = \frac{6}{12}$, $\frac{1}{3} = \frac{4}{12}$, so …

The key here is conditional probability. The formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ tells us the probability of $A$ happening given that $B$ has already occurred. It shrinks the sample space to just $B$, so we divide by $P(B)$.

We are given $P(A \mid B) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$. That lets us find $P(A \cap B)$ directly — the overlap of the two events. Once we have that, we can find $P(A \cup B)$ using the addition rule, and then $P(A' \cap B')$ is simply the complement of that union (De Morgan’s law).

Let’s go step by step.

1. Find $P(A \cap B)$ using conditional probability. From $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, we get

2. Find $P(A \cup B)$ using the addition rule. The formula is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Substitute the known values:

Get a common denominator (12):

3. Use De Morgan’s law to find $P(A' \cap B')$. $A' \cap B'$ is the complement of $A \cup B$. So $P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{3}{4} = \frac{1}{4}.$ …

Method: Multiplication theorem, then De Morgan for "neither"

Use this when you are given a conditional probability, must first recover the overlap, and finally want $P(A'\cap B')$ (the chance that neither event happens).

Steps

Step 1: Recover the intersection with the multiplication theorem.

This rearrangement of the conditional-probability definition turns a "given that" into a joint probability.

Step 2: Build the union. …

Common Mistakes

Mistake 1: Treating $P(A\mid B)$ as if it were $P(A\cap B)$.

Why it's wrong: the conditional already has $P(B)$ divided out. Correct approach: multiply back, $P(A\cap B)=P(A\mid B)\,P(B)$.

Mistake 2: Confusing "neither" with "not both." …