Q.If $P(A) = \dfrac{2}{5}$, $P(B) = \dfrac{3}{10}$ and $P(A \cap B) = \dfrac{1}{5}$, then $P(A' \mid B') \cdot P(B' \mid A')$ is equal to
(A) $\dfrac{5}{6}$
(B) $\dfrac{5}{7}$
(C) $\dfrac{25}{42}$
(D) $1$

Concept: Conditional Probability

We need $P(A' \mid B') \cdot P(B' \mid A')$.

By definition, $P(A' \mid B') = \frac{P(A' \cap B')}{P(B')}$ and $P(B' \mid A') = \frac{P(A' \cap B')}{P(A')}$.

So the product is $\frac{[P(A' \cap B')]^2}{P(A') \cdot P(B')}$.

Step 1: Find $P(A')$ and $P(B')$

$P(A') = 1 - \frac{2}{5} = \frac{3}{5}$,

$P(B') = 1 - \frac{3}{10} = \frac{7}{10}$.

Step 2: Find $P(A' \cap B')$ using De Morgan’s law

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

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{5} + \frac{3}{10} - \frac{1}{5} = \frac{4}{10} + \frac{3}{10} - \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$. …

We are given $P(A) = \frac{2}{5}$, $P(B) = \frac{3}{10}$, and $P(A \cap B) = \frac{1}{5}$. We need $P(A' \mid B') \cdot P(B' \mid A')$.

The key idea: conditional probability measures the chance of one event given that another has occurred. Here, both conditions involve complements — so we first find probabilities of the complements and their intersection.

Step 1: Find $P(A')$ and $P(B')$.

Since $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$:

Step 2: Find $P(A' \cap B')$.

By De Morgan’s law, $A' \cap B' = (A \cup B)'$. So $P(A' \cap B') = 1 - P(A \cup B)$.

We need $P(A \cup B)$ first. Using the addition rule:

Convert to tenths: $\frac{2}{5} = \frac{4}{10}$, $\frac{1}{5} = \frac{2}{10}$. So:

Thus:

Step 3: Write the conditional probabilities. …

Method: Conditional probabilities of complements

Use this when the events being conditioned on are complements ($A'$, $B'$) and you must combine two such conditionals.

Steps

Step 1: Get the complement marginals.

Step 2: Get the joint of the complements with De Morgan's law.

$A'\cap B'=(A\cup B)'$, so

$P(A'\cap B')=1-P(A\cup B),\quad\text{with}\quad P(A\cup B)=P(A)+P(B)-P(A\cap B).$ …

Common Mistakes

Mistake 1: Using $P(A'\mid B')=1-P(A\mid B)$.

Why it's wrong: conditional probability does not distribute over complements, and here the condition also changes from $B$ to $B'$. Correct approach: go back to the definition, $P(A'\mid B')=\dfrac{P(A'\cap B')}{P(B')}$.

Mistake 2: Mishandling the joint of complements. …