Q.If $A$ and $B$ are such events that $P(A) > 0$ and $P(B) \neq 1$, then $P(A' \mid B')$ equals
(A) $1 - P(A \mid B)$
(B) $1 - P(A' \mid B)$
(C) $\dfrac{1 - P(A \cup B)}{P(B')}$
(D) $\dfrac{P(A')}{P(B')}$

Concept: Probability Complement Rule and conditional probability.

We start with the definition of conditional probability:

Since $A' \cap B' = (A \cup B)'$, we have:

$P(A' \mid B') = \frac{P((A \cup B)')}{P(B')} = \frac{1 - P(A \cup B)}{P(B')}$ …

Why the Complement Rule is the Right Lens

When you see $P(A' \mid B')$, your first instinct might be to reach for $1 - P(A \mid B)$ — but that’s a trap. Conditional probability doesn’t distribute over complements the way unconditional probability does. The complement rule for conditional probability is:

not $1 - P(A \mid B)$. That subtle shift in the condition is everything.

So the cleanest path is to start from the definition of conditional probability, rewrite the numerator using set algebra, and simplify.

Step-by-Step Derivation

1. Write the definition.

For any two events $X$ and $Y$ with $P(Y) > 0$:

Here $X = A'$ and $Y = B'$, so:

2. Express the intersection using De Morgan’s law.

$A' \cap B'$ is the complement of $A \cup B$:

Therefore:

3. Substitute back.

That’s exactly option (C).

4. Check why the other options fail. …

Method: Deriving a conditional-probability identity

For "which expression equals $P(A'\mid B')$" questions, derive the identity from first principles rather than guessing.

Steps

Step 1: Start from the definition.

Step 2: Rewrite the numerator with De Morgan.

$A'\cap B'=(A\cup B)'$, so $P(A'\cap B')=1-P(A\cup B)$.

Step 3: Assemble and reject the traps.

$P(A'\mid B')=\frac{1-P(A\cup B)}{P(B')}.$ …

Common Mistakes

Mistake 1: Choosing $1-P(A\mid B)$.

Why it's wrong: the complement rule for conditionals keeps the condition fixed — $P(A'\mid C)=1-P(A\mid C)$ — but here the condition is $B'$, not $B$. Correct approach: derive from the definition with condition $B'$.

Mistake 2: Choosing $\dfrac{P(A')}{P(B')}$. …