Ten persons with badges numbered 1 to 10 are | vedclass

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(C) Total number of ways to select $3$ persons out of $10$ is given by $n(S) = {}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. I

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$\frac{1}{12}$

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(C) Total number of ways to select $3$ persons out of $10$ is given by $n(S) = {}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. If the smallest badge number among the three selected is $5$, then $5$ must be one of the selected persons. The other two persons must be selected from the set of numbers greater than $5$, which is $\{6, 7, 8, 9, 10\}$. There are $5$ such numbers. Therefore, the number of ways to select the other two persons is $n(A) = {}^{5}C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$. The required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{10}{120} = \frac{1}{12}$.

Ten persons with badges numbered $1$ to $10$ are in a room. If three of them are asked to leave the room, then the probability that the person with the smallest badge number among the three who left is $5$, is:

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