If a family has 2 children, the sample space can be reprsented as S = {(b,b), (b,g), (g,b), (g,g)}. Thus the total number of outcomes is 4.

The total number of outcomes for the event that both children are girls is 1 which is E = {g,g} $\rightarrow P(E) = \large \frac{1}{4}$.

Case a) If the youngest is a girl, then the total number of favourable outcomes for this event is 2, which are F = {(b,g), (g,g)} $\rightarrow P(F) = \large \frac{2}{4} = \frac{1}{2}$.

$\Rightarrow E \cap F = {g,g} \rightarrow P (E \cap F) = \large \frac{1}{4}$

Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

$\Rightarrow P(E/F) = \Large \frac{\frac{1}{4}}{\frac{1}{2}}$ = $\large \frac{1}{2}$

Case b) If atleast one of the children is a girl, then the total number of favorable outcomes is 3, F = {(b,g), (g,b), (g,g)} $\rightarrow P(E) = \large \frac{3}{4}$

$\Rightarrow E \cap F = {g,g} \rightarrow P (E \cap F) = \large \frac{1}{4}$

$\Rightarrow P(E/F) = \Large \frac{\frac{1}{4}}{\frac{3}{4}}$ = $\large \frac{1}{3}$