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

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Solution

## $\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}S=\left\{BBB,BBG,BGB,BGG,GGG,GBG,GGB,GBB\right\},\mathrm{where}\mathrm{the}\mathrm{first}\mathrm{letter}\mathrm{in}\mathrm{each}\mathrm{element}\mathrm{represents}\mathrm{the}\mathrm{eldest}\mathrm{child}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{family}\mathrm{with}\mathrm{a}\mathrm{girl}\mathrm{as}\mathrm{the}\mathrm{eldest}\mathrm{child}\mathrm{and}\phantom{\rule{0ex}{0ex}}B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{family}\mathrm{with}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{girl}\mathrm{child}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},A=\left\{GGG,GBG,GGB,GBB\right\}\mathrm{and}B=\left\{BBG,BGB,BGG,GGG,GBG,GGB,GBB\right\}\phantom{\rule{0ex}{0ex}}⇒\mathrm{n}\left(A\right)=4,\mathrm{n}\left(B\right)=7\mathrm{and}\mathrm{n}\left(A\cap B\right)=\mathrm{n}\left(A\right)=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(A|B\right)=\frac{\mathrm{n}\left(A\cap B\right)}{\mathrm{n}\left(B\right)}=\frac{4}{7}$ Hence, the correct alternative is option (d).

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