<p>A box contains 100 tickets numbered 1, 2, … 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability is</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(min=5 | max≤10) = P(min=5 AND max≤10) / P(max≤10). The numerator requires one ticket to be 5 and the other to be in {6,7,8,9,10}, while the denominator counts all pairs with both tickets in {1,2,...,10}.
<p><strong>Step 1:</strong> Find P(max ≤ 10). When choosing 2 tickets from {1,2,...,10}, the number of ways = C(10,2) = 45. Total ways to choose 2 from 100 = C(100,2) = 4950. So P(max ≤ 10) = 45/4950.</p><p><strong>Step 2:</strong> Find P(min = 5 AND max ≤ 10). For minimum to be 5 and maximum ≤ 10, one ticket must be 5 and the other must be from {6,7,8,9,10}. Number of such pairs = 5. So P(min=5 AND max≤10) = 5/4950.</p><p><strong>Step 3:</strong> Apply conditional probability formula: P(min=5 | max≤10) = [P(min=5 AND max≤10)] / [P(max≤10)] = (5/4950) / (45/4950) = 5/45 = 1/9.</p><p><strong>∴ Answer: 1/9 (Option D)</strong></p>
Correct Answer: D