<p>Two integers are selected at random from the set \(\{1, 2, \ldots, 11\}\). Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is:</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(both even | sum even) = P(both even AND sum even) / P(sum even). Recognize that if both numbers are even, their sum is automatically even, so the intersection equals P(both even).
<p><strong>Step 1:</strong> Identify the set and partition it by parity.</p><p>Set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}</p><p>Even numbers: {2, 4, 6, 8, 10} — 5 numbers</p><p>Odd numbers: {1, 3, 5, 7, 9, 11} — 6 numbers</p><p><strong>Step 2:</strong> Count pairs with even sum.</p><p>Sum is even when: both even OR both odd</p><p>Both even: C(5,2) = 10 pairs</p><p>Both odd: C(6,2) = 15 pairs</p><p>Total pairs with even sum = 10 + 15 = 25</p><p><strong>Step 3:</strong> Apply conditional probability formula.</p><p>P(both even | sum even) = P(both even AND sum even) / P(sum even)</p><p>Since both even automatically gives even sum: P(both even AND sum even) = 10/C(11,2) = 10/55</p><p>P(sum even) = 25/55</p><p><strong>Step 4:</strong> Calculate the conditional probability.</p><p>P(both even | sum even) = (10/55) ÷ (25/55) = 10/25 = 2/5</p><p>∴ Answer: D (2/5)</p>
Correct Answer: D