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Step-by-Step Solution
Step 1: Identify the Events
Let us define two events:
A: The event that the minimum of the chosen numbers is 3.
B: The event that the maximum of the chosen numbers is 6.
We want to find the conditional probability $P(A \mid B)$, which by definition of conditional probability is:
$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$
Step 2: Total Number of Ways to Choose 3 Numbers
We are choosing 3 numbers from the set $ \{1, 2, 3, 4, 5, 6, 7, 8\} $ without replacement. The total number of ways to do this is:
$\binom{8}{3} = 56.$
Step 3: Calculate $P(B)$ (Probability that the Maximum is 6)
To have a maximum of 6, we must include 6 in our selection and the other two numbers must come from $ \{1, 2, 3, 4, 5\} $. The number of ways is:
$\binom{5}{2} = 10.$
Hence,
$P(B) = \frac{10}{56} = \frac{5}{28}.$
Step 4: Calculate $P(A \cap B)$ (Probability that the Minimum is 3 and the Maximum is 6)
For the minimum to be 3 and the maximum to be 6, we must include both 3 and 6 in our selection. The remaining one number must be chosen from $ \{4, 5\} $. The number of ways to choose this middle number is:
$\binom{2}{1} = 2.$
Therefore,
$P(A \cap B) = \frac{2}{56} = \frac{1}{28}.$
Step 5: Apply the Conditional Probability Formula
Using $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, we get:
$P(A \mid B) = \frac{\frac{1}{28}}{\frac{5}{28}} = \frac{1}{5}.$
Therefore, the probability that their minimum is 3, given that their maximum is 6, is
$\displaystyle \frac{1}{5}.$
