Assume that in a family, each child is equally likely to be a boy or…

The correct answer is Option (4) → $\frac{4}{7}$ ##

Here, $S = \{(B, B, B), (G, G, G), (B, G, G), (G, B, G), (G, G, B), (G, B, B), (B, G, B), (B, B, G)\}$ i.e., $n(S) = 8$

Let $E_1 = \text{Event that a family has atleast one girl, then}$

$E_1 = \{(G, B, B), (B, G, B), (B, B, G), (G, G, B), (B, G, G), (G, B, G), (G, G, G)\}$

i.e., $n(E_1) = 7$

$E_2 = \text{Event that the eldest child is a girl, then}$

$E_2 = \{(G, B, B), (G, G, B), (G, B, G), (G, G, G)\}$

$∴E_1 \cap E_2 = \{(G, B, B), (G, G, B), (G, B, G), (G, G, G)\}$ i.e., $n(E_1 \cap E_2) = 4$

$∴P(E_2 \mid E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)} = \frac{4/8}{7/8} = \frac{4}{7}$