Question

A fair die is rolled. Consider events $E=\{1,3,5\},F=\{2,3\}$ and $G=\{2,3,4,5\}$ Find
(i) $P\left(E|F\right)$ and $P\left(F|E\right)$
(ii) $P\left(E|G\right)$ and $P\left(G|E\right)$
(iii) $P\left((E\cup F)|G\right)$ and $P\left((E\cap F)|G\right)$

Answer 1

When a fair die is rolled, the sample space $S$ will be
$S=\{1,2,3,4,5,6\}$

It is given that $E=\{1,3,5\},F=\{2,3\}$ and $G=\{2,3,4,5\}$

$\therefore P\left(E\right)=\frac{3}{6}=\frac{1}{2}$

$P\left(F\right)=\frac{2}{6}=\frac{1}{3}$

$P\left(G\right)=\frac{4}{6}=\frac{2}{3}$

$\left(i\right)$ $E\cap F=\left\{3\right\}$

$\therefore P(E\cap F)=\frac{1}{6}$

$\therefore P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}$

$P\left(F|E\right)=\frac{P(E\cap F)}{\left(E\right)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}$

$\left(ii\right)$ $E\cap G=\{3,5\}$

$\therefore P(E\cap G)=\frac{2}{6}=\frac{1}{3}$

$\therefore P\left(E|G\right)=\frac{P(E\cap G)}{P\left(G\right)}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$

$P\left(G|E\right)=\frac{P(E\cap G)}{P\left(E\right)}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

$\left(iii\right)$ $E\cup F=\{1,2,3,5\}$

$(E\cup F)\cap G=\{1,2,3,5\}\cap \{2,3,4,5\}=\{2,3,5\}$

$E\cap F=\left\{3\right\}$

$(E\cap F)\cap G=\left\{3\right\}\cap \{2,3,4,5\}=\left\{3\right\}$

$\therefore P(E\cup G)=\frac{4}{6}=\frac{2}{3}$

$P((E\cup F)\cap G)=\frac{3}{6}=\frac{1}{2}$

$P(E\cap F)=\frac{1}{6}$

$P((E\cap F)\cap G)=\frac{1}{6}$

$\therefore P\left((E\cup F)|G\right)=\frac{P((E\cup F)\cap G)}{P\left(G\right)}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{1}{2}\times \frac{3}{2}=\frac{3}{4}$

$P\left((E\cap F)|G\right)=\frac{P((E\cap G)\cap G)}{P\left(G\right)}=\frac{\frac{1}{6}}{\frac{2}{3}}=\frac{1}{6}\times \frac{3}{2}=\frac{1}{4}$