<p>Three numbers are chosen at random without replacement from \(\{1, 2, 3, \ldots, 8\}\). The probability that their minimum is 3, given that their maximum is 6, is:</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(min=3|max=6) = P(min=3 AND max=6)/P(max=6). The numerator requires all three numbers in {3,4,5,6} with 3 and 6 both present, while denominator counts all triples with max=6.
<p><strong>Step 1:</strong> Find P(max = 6). We need triples from {1,2,3,4,5,6} with at least one 6.</p><p>Total ways to choose 3 from {1,2,...,8}: C(8,3) = 56</p><p>Triples with max = 6: Choose 3 from {1,2,3,4,5,6} with at least one being 6 = C(6,3) - C(5,3) = 20 - 10 = 10</p><p><strong>Step 2:</strong> Find P(min = 3 AND max = 6). We need triples with minimum 3 and maximum 6.</p><p>Both 3 and 6 must be selected. Third number from {4,5}: C(2,1) = 2</p><p>These triples are: {3,4,6} and {3,5,6}</p><p><strong>Step 3:</strong> Apply conditional probability.</p><p>P(min=3|max=6) = (Number of favorable outcomes)/(Number of outcomes with max=6) = 2/10 = 1/5</p><p>∴ Answer: B</p>
Correct Answer: B