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(A) Let $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$. We choose $3$ numbers from $S$ without replacement.Let $A$ be the event that the maximum of the chosen numbe

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$\frac{1}{5}$

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(A) Let $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$. We choose $3$ numbers from $S$ without replacement.Let $A$ be the event that the maximum of the chosen numbers is $6$.Let $B$ be the event that the minimum of the chosen numbers is $3$.We want to find the conditional probability $P(B|A) = \frac{P(A \cap B)}{P(A)}$.For event $A$ (maximum is $6$): The numbers must be chosen from $\{1, 2, 3, 4, 5, 6\}$ such that $6$ is included. The other $2$ numbers must be chosen from $\{1, 2, 3, 4, 5\}$.Number of ways for $A = \binom{5}{2} = 10$.For event $A \cap B$ (maximum is $6$ and minimum is $3$): The numbers must be chosen from $\{3, 4, 5, 6\}$ such that $3$ and $6$ are included. The remaining $1$ number must be chosen from $\{4, 5\}$.Number of ways for $A \cap B = \binom{2}{1} = 2$.Thus,$P(B|A) = \frac{n(A \cap B)}{n(A)} = \frac{2}{10} = \frac{1}{5}$.

Three numbers are chosen at random without replacement from $\{1, 2, 3, 4, 5, 6, 7, 8\}$. The probability that their minimum is $3$,given that their maximum is $6$,is:

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