$S = \{ (B,b) , (B,g), (G,b) , (G,g)\}$

B =elder male child,

b = younger male child,

G= elder female child,

g = younger female child

(a) Let E_1 be the event that the youngest is a girl . $E_1 = \{ (Bg) , (Gg)\}$

$P(E_1)= \frac{2}{4} = \frac{1}{2}$

Let $E_2$ be the event that both are girls

$E_2 = \{(Gg)\} = P (E_2)= \frac{1}{4}$

$\therefore E_2 \cap E_1 = \{Gg\} . P(E_2 \cap E_1) = \frac{1}{4}$

$\begin{align*}\therefore P (\frac{E_2}{E_1} ) &= \frac{P (E_2 \cap E_1)}{P(E_1)} \\ & = \frac{1/4}{1/2} \\ & = 1/2 \end{align*}$

(b) Let $E_3 $ be the event that at least one is a girl = $\{(Bg), (Gb), (Gg)\}$

$P(E_3) = 3/4$

$P(E_3 \cap E_2) = \frac{1}{4}$

$\therefore \frac{P(E \cap E_2)}{P(E_3)} = \frac{1/4}{3/4} = 1/3$

Pre-natal determination is a crime. Hence we can make them aware of this crime and educate them not to indulge in such activities.