1. Key Concepts and Formulas

   • Conditional Probability: The probability of an event $A$ occurring given that another event $B$ has already occurred is denoted as $P(A|B)$. The fundamental formula is $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
   • Reduced Sample Space Method: For problems involving counting outcomes (like combinations), it is often more efficient to calculate conditional probability using the number of outcomes: $P(A|B) = \frac{\text{Number of outcomes where } A \text{ and } B \text{ both occur}}{\text{Number of outcomes where } B \text{ occurs}} = \frac{N(A \cap B)}{N(B)}$ This method avoids calculating the total number of possible outcomes in the universal set, as it cancels out.
   • Combinations: The number of ways to choose $k$ distinct items from a set of $n$ distinct items without regard to the order is given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. This is used for "without replacement" selection.

2. Step-by-Step Solution

   Step 1: Understand the Problem and Define Events We are choosing three distinct numbers at random without replacement from the set $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$. Let's define the events:

   • Event $A$: The minimum of the three chosen numbers is $3$.
   • Event $B$: The maximum of the three chosen numbers is $6$. Our goal is to calculate $P(A|B)$, which is the probability that the minimum is $3$, given that the maximum is $6$.

   Step 2: Determine the Size of the Reduced Sample Space ($N(B)$) We first need to find the number of ways to choose $3$ numbers such that their maximum is $6$. This forms our reduced sample space for event $B$.

   • Condition for "maximum is 6":
     1. The number $6$ must be one of the three chosen numbers.
     2. The other two numbers must be chosen from the set of numbers strictly smaller than $6$. This ensures that $6$ is indeed the largest among the three.
   • Identifying the pool for the other numbers: The numbers in $S$ that are strictly smaller than $6$ are $\{1, 2, 3, 4, 5\}$. There are $5$ such numbers.
   • Calculating $N(B)$: We choose $6$ in $\binom{1}{1}$ way. We choose $2$ additional numbers from the $5$ available numbers in $\{1, 2, 3, 4, 5\}$ in $\binom{5}{2}$ ways. $N(B) = \binom{1}{1} \times \binom{5}{2} = 1 \times \frac{5 \times 4}{2 \times 1} = 1 \times 10 = 10$ The $10$ sets of numbers where the maximum is $6$ are: $\{1,2,6\}, \{1,3,6\}, \{1,4,6\}, \{1,5,6\}, \{2,3,6\}, \{2,4,6\}, \{2,5,6\}, \{3,4,6\}, \{3,5,6\}, \{4,5,6\}$.

   Step 3: Determine the Number of Outcomes for the Intersection ($N(A \cap B)$) Next, we need to find the number of ways to choose $3$ numbers such that their minimum is $3$ AND their maximum is $6$. This is the intersection of events $A$ and $B$.

   • Condition for "minimum is 3 AND maximum is 6":
     1. The number $3$ must be one of the chosen numbers (as it's the minimum).
     2. The number $6$ must be one of the chosen numbers (as it's the maximum).
     3. The third number must be strictly greater than $3$ (to maintain $3$ as the minimum) AND strictly less than $6$ (to maintain $6$ as the maximum).
   • Identifying the pool for the third number: The numbers in $S$ that are strictly between $3$ and $6$ are $\{4, 5\}$. There are $2$ such numbers.
   • Calculating $N(A \cap B)$: We choose $3$ in $\binom{1}{1}$ way. We choose $6$ in $\binom{1}{1}$ way. We choose $1$ additional number from the $2$ available numbers in $\{4, 5\}$ in $\binom{2}{1}$ ways. $N(A \cap B) = \binom{1}{1} \times \binom{1}{1} \times \binom{2}{1} = 1 \times 1 \times 2 = 2$ The $2$ sets of numbers where the minimum is $3$ and the maximum is $6$ are: $\{3, 4, 6\}$ and $\{3, 5, 6\}$.

   Step 4: Calculate the Conditional Probability Using the reduced sample space formula, we can now find $P(A|B)$: $P(A|B) = \frac{N(A \cap B)}{N(B)}$ Substitute the values calculated in Step 2 and Step 3: $P(A|B) = \frac{2}{10} = \frac{1}{5}$

3. Common Mistakes & Tips

   • Misinterpreting "Minimum" or "Maximum": Always remember that if $k$ is the minimum, $k$ must be included, and all other numbers must be strictly greater than $k$. Similarly for the maximum.
   • Incorrect Range for Intermediate Numbers: When both minimum ($m$) and maximum ($M$) are specified, the remaining numbers must be chosen from the set of integers strictly between $m$ and $M$. E.g., for min 3 and max 6, the intermediate numbers are from $\{4,5\}$.
   • Forgetting "Without Replacement": This implies that numbers must be distinct, which is correctly handled by using combinations ($\binom{n}{k}$).
   • Not Using Reduced Sample Space: While the full conditional probability formula ($P(A|B) = \frac{P(A \cap B)}{P(B)}$) works, using counts directly ($P(A|B) = \frac{N(A \cap B)}{N(B)}$) is generally simpler for counting problems as the total sample space size cancels out.

4. Summary To find the probability that the minimum of three chosen numbers is $3$, given that their maximum is $6$, we utilized the conditional probability concept with a reduced sample space. We first identified all possible sets of three numbers whose maximum is $6$, finding $10$ such sets. Then, from these $10$ sets, we identified the subset where the minimum is also $3$, finding $2$ such sets. The conditional probability is the ratio of these counts, which is $\frac{2}{10}$, simplifying to $\frac{1}{5}$.

The final answer is \boxed{\text{1 \over 5}}, which corresponds to option (B).