A couple has two children,

If a couple has two children, then the sample space is

S = {(b, b), (b, g), (g, b), (g, g)}

(i) Let E and F respectively denote the events that both children are males and at least one of the children is a male.

$\therefore \mathrm{E} \cap \mathrm{F}=\{(b, b)\} \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{4}$

$P(E)=\frac{1}{4}$

$P(F)=\frac{3}{4}$

$\Rightarrow P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

$\Rightarrow P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

(ii) Let A and B respectively denote the events that both children are females and the elder child is a female.

$\mathrm{A}=\{(g, g)\} \Rightarrow \mathrm{P}(\mathrm{A})=\frac{1}{4}$

$\mathrm{~B}=\{(g, b),(g, g)\} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{2}{4}$

$\mathrm{~A} \cap \mathrm{B}=\{(g, g)\} \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$

$\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{1}{\frac{4}{2}}=\frac{1}{2}$