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Three numbers are chosen at random without replacement from {1, 2, 3, ...... 8}. The probability that their minimum is 3, given that their maximum is 6, is

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\begin{array}{l} \mathrm{A} \rightarrow \text { Maximum is } 6 \\ \mathrm{B} \rightarrow \text { Minimum is } 3 \\ \mathrm{P}(\mathrm{A})=\frac{{ }^{5} \mathrm{C}_{2}}{{ }^{8} \mathrm{C}_{3}} \\ \mathrm{P}(\mathrm{B})=\frac{{ }^{5} \mathrm{C}_{2}}{{ }^{8} \mathrm{C}_{3}} \\ \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{{ }^{2} \mathrm{C}_{1}}{{ }^{8} \mathrm{C}_{3}}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{{ }^{2} \mathrm{C}_{1}}{{ }^{5} \mathrm{C}_{2}}=\frac{2}{10}=\frac{1}{5} \\ \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{1}{2} \end{array}

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Satyajeet Kumar

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