If $A$ and $B$ are two events such that $P(A)=1 / 4, P(B)=1 / 2$ and $P(A \cap B)=1 / 8$, find $P($ not $A$ and not B).

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If $A$ and $B$ are two events such that $P(A)=1 / 4, P(B)=1 / 2$ and $P(A \cap B)=1 / 8$ , find $P($ not $A$ and not B).

If $A$ and $B$ are two events such that $P(A)=1 / 4, P(B)=1 / 2$ and $P(A \cap B)=1 / 8$ , find $P($ not $A$ and not B).

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.

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