Sample space: S = {MM, MF, FM, FF}, where M = male and F = female.

(i) To find P (both children are males, if it is known that at least one of the children is male).

A: Event that both children are male, and B: event that at least one of them is a male.

A: {MM} and B: {MF, FM, MM} $\rightarrow$ P (A $\cap$ B) = {MM}

Probability that both are males, if we know one is a female = $P \large(\frac{A}{B}) =$$ \large (\frac{P(A\;\cap\; B)}{P(B)})$

Given S = {MM, MF, FM, FF}, we can see that: P (A) = $\large\frac{1}{4}$; P (B) = $\large\frac{3}{4}$; P (A $\cap$ B) = $\large\frac{1}{4}$

Therefore, $P \large(\frac{B}{A}) =$$ \large (\frac{P(A\;\cap\; B)}{P(A)}) = \large\frac{\frac{1}{4}}{\frac{3}{4} }= \frac{1}{3}$

(ii) To find the probability that both are females, if we know that the elder child is female.

Let A: event that both are females and B: event that elder one is a female child.

A: {FF} and B: {FF, FF} $\rightarrow$ P (A $\cap$ B) = {FF}.

P (both are females given the elder child is female) = $P \large(\frac{A}{B}) =$$ \large (\frac{P(A\;\cap\; B)}{P(B)})$

Given S = {MM, MF, FM, FF}, we can see that: P (A) = $\large\frac{1}{4}$; P (B) = $\large\frac{2}{4} = \frac{1}{2}$; P (A $\cap$ B) = $\large\frac{1}{4}$

Therefore, $P \large(\frac{B}{A}) =$$ \large (\frac{P(A\;\cap\; B)}{P(B)}) = \large\frac{\frac{1}{4}}{\frac{1}{2} }= \frac{1}{2}$