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(A) The total number of ways to choose $3$ numbers from $10$ is $n(S) = {}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. Let $A$

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$\frac{11}{40}$

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(A) The total number of ways to choose $3$ numbers from $10$ is $n(S) = {}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. Let $A$ be the event that the minimum of the chosen numbers is $3$. This means $3$ is chosen,and the other two numbers must be chosen from $\{4, 5, 6, 7, 8, 9, 10\}$. Thus,$n(A) = {}^{7}C_2 = 21$. Let $B$ be the event that the maximum of the chosen numbers is $7$. This means $7$ is chosen,and the other two numbers must be chosen from $\{1, 2, 3, 4, 5, 6\}$. Thus,$n(B) = {}^{6}C_2 = 15$. Let $A \cap B$ be the event that the minimum is $3$ $AND$ the maximum is $7$. This means $3$ and $7$ are chosen,and the third number must be chosen from $\{4, 5, 6\}$. Thus,$n(A \cap B) = {}^{3}C_1 = 3$. Using the inclusion-exclusion principle,$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 21 + 15 - 3 = 33$. The required probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{33}{120} = \frac{11}{40}$.

If three numbers are drawn at random successively without replacement from a set $S = \{1, 2, \ldots, 10\}$,then the probability that the minimum of the chosen numbers is $3$ or their maximum is $7$ is:

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