Q.In Exercise 64 above, $P(B \mid A')$ is equal to
(A) $\dfrac{1}{5}$
(B) $\dfrac{3}{10}$
(C) $\dfrac{1}{2}$
(D) $\dfrac{3}{5}$

From Exercise 64: $P(B)=\tfrac35,\ P(A\mid B)=\tfrac12,\ P(A\cup B)=\tfrac45.$

$P(A\cap B)=P(A\mid B)\,P(B)=\tfrac12\cdot\tfrac35=\tfrac{3}{10}.$

$P(A)=P(A\cup B)-P(B)+P(A\cap B)=\tfrac45-\tfrac35+\tfrac{3}{10}=\tfrac12,$ so $P(A')=\tfrac12.$ …

Collect the data from Exercise 64

Step 1 — $P(A\cap B)$

By the multiplication rule, $P(A\cap B)=P(A\mid B)\,P(B)=\tfrac12\cdot\tfrac35=\tfrac{3}{10}.$

Step 2 — $P(A)$ and $P(A')$

From the addition rule, $P(A)=P(A\cup B)-P(B)+P(A\cap B)=\tfrac45-\tfrac35+\tfrac{3}{10}=\tfrac{2}{10}+\tfrac{3}{10}=\tfrac12.$

Hence $P(A')=1-\tfrac12=\tfrac12.$

Step 3 — $P(B\cap A')$ …

Method: Conditioning on a complement event

Use this to compute $P(B\mid A')$ — a probability conditioned on an event not happening.

Steps

Step 1: Rebuild the pieces you need.

From the given data recover $P(A\cap B)=P(A\mid B)P(B)$, then $P(A)$ from the addition rule, and hence $P(A')=1-P(A)$.

Step 2: Find the joint with the complement.

The event $B$ splits into the part inside $A$ and the part outside:

$P(B\cap A')=P(B)-P(A\cap B).$ …

Common Mistakes

Mistake 1: Dividing by $P(A)$ instead of $P(A')$.

Why it's wrong: the condition here is $A'$ ("$A$ did not occur"), so its probability is the denominator. Correct approach: $P(B\mid A')=\dfrac{P(B\cap A')}{P(A')}$.

Mistake 2: Computing $P(B\cap A')$ incorrectly. …