Question

If E and F are events such that $P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{2}$ and $P(E\cap F)=\frac{1}{8}$ find $P(E^{\prime }\cap F^{\prime })$

Answer 1

$P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{2},P(E\cap F)=\frac{1}{8}$

$P(E^{\prime }\cap F^{\prime })=P(E\cup F{)}^{\prime }$ [By DE Morgan law]

$=1-P(E\cup F)$

$\therefore P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)$

$=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$

$=\frac{5}{8}$

$\therefore P(E^{\prime }\cap F^{\prime })=1-\frac{5}{8}$

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

$=\frac{3}{8}$.