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

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It is given that, $\mathrm{P}(\mathrm{A}) \frac{1}{4}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$

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

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

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

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

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

$=\frac{3}{8}$

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