Question

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, then what is the constitutional probability that both are girls? Given that

(i) the youngest is a girl (b) at least one is a girl. [CBSE 2014]

Answer 1

Let G and B represents respectively a girl and a boy.

As, the possible outcomes of a family having two children is given by the set, S = {GG, GB, BG, BB}, where first letter represents the elder child.

So, n(S) = 4

Now,

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}: \\ A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{two}\mathrm{girl}\mathrm{children}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{a}\mathrm{girl}\mathrm{as}\mathrm{the}\mathrm{youngest}\mathrm{child}. \\ A=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\right)=1, \\ B=\left\{GG,BG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(B\right)=2\mathrm{and} \\ A\cap B=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\cap B\right)=1 \\ \mathrm{So},\mathrm{P}\left(A|B\right)=\frac{\mathrm{n}\left(A\cap B\right)}{\mathrm{n}\left(B\right)}=\frac{1}{2} \\ \left(\mathrm{ii}\right)\mathrm{Let}: \\ A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{two}\mathrm{girl}\mathrm{children}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{girl}\mathrm{child}. \\ A=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\right)=1, \\ B=\left\{GG,BG,GB\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(B\right)=3\mathrm{and} \\ A\cap B=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\cap B\right)=1 \\ \mathrm{So},\mathrm{P}\left(A|B\right)=\frac{\mathrm{n}\left(A\cap B\right)}{\mathrm{n}\left(B\right)}=\frac{1}{3} \end{gathered}$$