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Question

# Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

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Solution

## Consider the given events. A = Both the children are girls. B = The youngest child is a girl. C = At least one child is a girl. $\mathrm{Clearly},\phantom{\rule{0ex}{0ex}}S=\left\{{B}_{1}{B}_{2},{B}_{1}{G}_{2},{G}_{1}{B}_{2},{G}_{1}{G}_{2}\right\}\phantom{\rule{0ex}{0ex}}A=\left\{{G}_{1}{G}_{2}\right\}\phantom{\rule{0ex}{0ex}}B=\left\{{B}_{1}{G}_{2},{G}_{1}{G}_{2}\right\}\phantom{\rule{0ex}{0ex}}C=\left\{{B}_{1}{G}_{2},{G}_{1}{B}_{2},{G}_{1}{G}_{2}\right\}$ $A\cap B=\left\{{G}_{1}{G}_{2}\right\}\mathrm{and}A\cap C=\left\{{G}_{1}{G}_{2}\right\}\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\mathrm{Required}\mathrm{probability}=P\left(A/B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{Required}\mathrm{probability}=P\left(A/C\right)=\frac{n\left(A\cap B\right)}{n\left(C\right)}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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