A fair die is rolled. Consider events

Solution:

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}

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

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

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

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

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

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

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

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

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

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

$\mathrm{P}(\mathrm{G} \mid \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{E})}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

(iii) E ∪ F = {1, 2, 3, 5}

(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}

E ∩ F = {3}

(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

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

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

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

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

$\therefore \mathrm{P}((\mathrm{E} \cup \mathrm{F}) \mid \mathrm{G})=\frac{\mathrm{P}((\mathrm{E} \cup \mathrm{F}) \cap \mathrm{G})}{\mathrm{P}(\mathrm{G})}$

$=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{1}{2} \times \frac{3}{2}=\frac{3}{4}$

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

$=\frac{\frac{1}{6}}{\frac{2}{3}}=\frac{1}{6} \times \frac{3}{2}=\frac{1}{4}$