If A and B are two events such that
Question:

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

Solution:

It is given that, $\mathrm{P}(\mathrm{A})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$

$P($ not on $A$ and not on $B)=P\left(A^{\prime} \cap B^{\prime}\right)$

$\mathrm{P}($ not on $\mathrm{A}$ and not on $\mathrm{B})=\mathrm{P}((\mathrm{A} \cup \mathrm{B}))^{\prime}\left[\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}=(\mathrm{A} \cup \mathrm{B})^{\prime}\right]$

$=1-P(A \cup B)$

$=1-[P(A)+P(B)-P(A \cap B)]$

$=1-\left[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right]$

$=1-\frac{5}{8}$

$=\frac{3}{8}$

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