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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]

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Solution

## 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, $\left(\mathrm{i}\right)\mathrm{Let}:\phantom{\rule{0ex}{0ex}}A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{two}\mathrm{girl}\mathrm{children}\mathrm{and}\phantom{\rule{0ex}{0ex}}B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{a}\mathrm{girl}\mathrm{as}\mathrm{the}\mathrm{youngest}\mathrm{child}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}A=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\right)=1,\phantom{\rule{0ex}{0ex}}B=\left\{GG,BG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(B\right)=2\mathrm{and}\phantom{\rule{0ex}{0ex}}A\cap B=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\cap B\right)=1\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{P}\left(A|B\right)=\frac{\mathrm{n}\left(A\cap B\right)}{\mathrm{n}\left(B\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{Let}:\phantom{\rule{0ex}{0ex}}A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{two}\mathrm{girl}\mathrm{children}\mathrm{and}\phantom{\rule{0ex}{0ex}}B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{the}\mathrm{family}\mathrm{has}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{girl}\mathrm{child}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}A=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\right)=1,\phantom{\rule{0ex}{0ex}}B=\left\{GG,BG,GB\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(B\right)=3\mathrm{and}\phantom{\rule{0ex}{0ex}}A\cap B=\left\{GG\right\}\mathrm{i}.\mathrm{e}.\mathrm{n}\left(A\cap B\right)=1\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{P}\left(A|B\right)=\frac{\mathrm{n}\left(A\cap B\right)}{\mathrm{n}\left(B\right)}=\frac{1}{3}$

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