Question

Mark the correct alternative in the following question:

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

$\left(\mathrm{a}\right)\frac{1}{2}\left(\mathrm{b}\right)\frac{1}{3}\left(\mathrm{c}\right)\frac{2}{3}\left(\mathrm{d}\right)\frac{4}{7}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ S=\left\{BBB,BBG,BGB,BGG,GGG,GBG,GGB,GBB\right\},\mathrm{where}\mathrm{the}\mathrm{first}\mathrm{letter}\mathrm{in}\mathrm{each}\mathrm{element}\mathrm{represents}\mathrm{the}\mathrm{eldest}\mathrm{child} \\ \mathrm{Let}: \\ A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{family}\mathrm{with}\mathrm{a}\mathrm{girl}\mathrm{as}\mathrm{the}\mathrm{eldest}\mathrm{child}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{family}\mathrm{with}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{girl}\mathrm{child} \\ \mathrm{So},A=\left\{GGG,GBG,GGB,GBB\right\}\mathrm{and}B=\left\{BBG,BGB,BGG,GGG,GBG,GGB,GBB\right\} \\ \Rightarrow \mathrm{n}\left(A\right)=4,\mathrm{n}\left(B\right)=7\mathrm{and}\mathrm{n}\left(A\cap B\right)=\mathrm{n}\left(A\right)=4 \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)=\frac{\mathrm{n}\left(A\cap B\right)}{\mathrm{n}\left(B\right)}=\frac{4}{7} \end{gathered}$$

Hence, the correct alternative is option (d).