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(B) The set $A$ is given by $A = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. The total number of elements in $A$ is $n(A) = 10$.The total number of ways to c

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$\frac{1}{40}$

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(B) The set $A$ is given by $A = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. The total number of elements in $A$ is $n(A) = 10$.The total number of ways to choose $3$ numbers from $10$ is $n(E) = {}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.For the minimum to be $3$ and the maximum to be $7$,the chosen set must contain $3$ and $7$. The third number must be chosen from the set $\{4, 5, 6\}$.Thus,the favorable outcomes are $\{3, 4, 7\}, \{3, 5, 7\}, \{3, 6, 7\}$.The number of favorable outcomes is $n(F) = 3$.The required probability is $P = \frac{n(F)}{n(E)} = \frac{3}{120} = \frac{1}{40}$.

Three numbers are chosen at random without replacement from the set $A = \{x \in \mathbb{Z} \mid 2 \leq x \leq 11\}$. The probability that the minimum of the chosen numbers is $3$ and the maximum is $7$ is

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