Three numbers are chosen at random from 1,2,3....10 without replacement. The probability that the minimum of the chosen numbers in 4 or the maximum is 8 is?

\frac{11}{40}

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\frac{4}{10}

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\frac{1}{12}

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\frac{3}{5}

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Solution

The correct option is A $\frac{11}{40}$
Probability of 4 being the minimum number of the three numbers selected = $\frac{^{6} C_{2}}{^{10} C_{3}}$
Probability of 8 being the maximum number of the three numbers selected = $\frac{^{7} C_{2}}{^{10} C_{3}}$
The probability of 4 being the minimum number and 8 being the maximum number = $\frac{3}{^{10} C_{3}}$
Therefore, required probability is $\frac{^{6} C_{2}}{^{10} C_{3}}$ + $\frac{^{7} C_{2}}{^{10} C_{3}}$ - $\frac{3}{^{10} C_{3}}$ = $\frac{11}{40}$

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