Question

If $E$ and $F$ are events such that $P\left(E\right)=$ $\frac{1}{4}$, $P\left(F\right)=$ $\frac{1}{2}$ and $P\left(EandF\right)=$ $\frac{1}{8}$ find: (i) P(E or F) (ii) P(not E and not F)

Answer 1

We have given, $P\left(E\right)=$ $\frac{1}{4}$, $P\left(F\right)=$ $\frac{1}{2}$ and P(E and F) = $\frac{1}{8}$
(i) We know that P(E or F) = P(E) + P(F) - P(E and F)
$\therefore$ P (E or F) = $\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}$
(ii) From (i), P(E or F) = $P(E\cup F)=\frac{5}{8}$
We have ${(E\cup F)}^{\prime }=(E^{\prime }\cap F^{\prime })$ [By De Morgan's law]
$\therefore P(E^{\prime }\cap F^{\prime })=P{(E\cup F)}^{\prime }$
Now $P{(E\cup F)}^{\prime }=1-P(E\cup F)=1-\frac{5}{8}=\frac{3}{8}$
$\therefore P(E^{\prime }\cap F^{\prime })=\frac{3}{8}$
Thus P(not E and not F) = $\frac{3}{8}$