A fair die is rolled. Consider events E = 1, 3, 5 , F= 2, 3 and G… - Vidyadip

Question

A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$
$\text{P}(\text{E}|\text{G})\ \text{and}\ \text{P}(\text{G}|\text{E})$

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Answer

$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \text{P}\left(\text{G}\right)=\frac{\text{n}\left(\text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{4}{6}$
$\text{E}\ \cap\ \text{G}=(3,\ 5)\ \Rightarrow\ \ \ \ \ \text{n}\left(\text{E}\cap\text{G}\right)=2$
$\text{P}\left(\text{E}\cap\text{G}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{P}\left(\text{E}|\text{G}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{G}\right)}{\text{P}\left(\text{G}\right)}=\frac{\frac{2}{6}}{\frac{4}{6}}=\frac{2}{4}=\frac{1}{2}\ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{G}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{G}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{2}{6}}{\frac{3}{6}}=\frac{2}{3}$

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