Question

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

Answer 1

$\text{Here P (E)}=\frac{1}{4},P\left(F\right)=\frac{1}{2}\text{and}P(E\cap F)=\frac{1}{8}\left(i\right)\text{We know that}P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}\text{(ii) P (not E and not F)}=P(\overline{E}\cap \overline{F})=P\overline{(E\cup F)}=1-P(E\cup F)=1-\frac{5}{8}=\frac{8-5}{8}=\frac{3}{8}$