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Question

Mother, father and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P (A/B) and P (B/A).

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Solution

Consider the given events. A = Son standing on one end B = Father standing in the middle $\mathrm{Clearly},\phantom{\rule{0ex}{0ex}}S=\left\{MFS,MSF,FSM,FMS,SMF,SFM\right\}\phantom{\rule{0ex}{0ex}}A=\left\{MFS,FMS,SMF,SFM\right\},\phantom{\rule{0ex}{0ex}}B=\left\{MFS,SFM\right\}$ $\mathrm{Now},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}A\cap B=\left\{MFS,SFM\right\}\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\mathrm{Required}\mathrm{probability}=P\left(A/B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}=\frac{2}{2}=1\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{Required}\mathrm{probability}=P\left(B/A\right)=\frac{n\left(A\cap B\right)}{n\left(A\right)}=\frac{2}{4}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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