Solution:
Given $P(A)=1 / 4, P(B)=1 / 2$ and $P(A \cap B)=1 / 8$
Concept: $P($ not $A$ and $\operatorname{not} B)=P\left(A^{\prime} \cap B^{\prime}\right)$
As, $\left{A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}\right}$
$\Rightarrow P(\operatorname{not} A$ and $\operatorname{not} B)=P\left((A \cup B)^{\prime}\right)$
$=1-P(A \cup B)$
$=1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]$
Substituting the value and evaluating we get,
$=1-\left[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right]$
$=1-\left[\frac{5}{8}\right]=\frac{3}{8}$
Thus the probability of event $P($ not $A$ and not B) is 3/8.