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Question

(C) Let $S$ be the set $\{1, 2, \ldots, 10\}$. The total number of ways to choose $3$ numbers is $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2

Vedclass · Accepted Answer

$\frac{11}{40}$

Vedclass · Answer

(C) Let $S$ be the set $\{1, 2, \ldots, 10\}$. The total number of ways to choose $3$ numbers is $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. Let $A$ be the event that the minimum is $3$. To have $3$ as the minimum, we must choose $3$ and two numbers from $\{4, 5, \ldots, 10\}$. The number of ways is $^7C_2 = 21$. Let $B$ be the event that the maximum is $7$. To have $7$ as the maximum, we must choose $7$ and two numbers from $\{1, 2, \ldots, 6\}$. The number of ways is $^6C_2 = 15$. The intersection $A \cap B$ is the event where the minimum is $3$ and the maximum is $7$. This means we choose $3$, $7$, and one number from $\{4, 5, 6\}$. The number of ways is $^3C_1 = 3$. By the inclusion-exclusion principle, the number of favorable outcomes is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 21 + 15 - 3 = 33$. The probability is $P(A \cup B) = \frac{33}{120} = \frac{11}{40}$.

Three numbers are chosen at random without replacement from $\{1, 2, \ldots, 10\}$. The probability that the minimum of the chosen numbers is $3$ or their maximum is $7$ is:

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