<p>Two integers are selected at random from \(\{1, 2, 3, \ldots, 11\}\). Given that the sum is even, the conditional probability that both numbers are odd is <em>[JEE Main 2019]</em></p>
Step-by-Step Solution
Key Concept: Sum is even iff both odd or both even. From {1,...,11}: 6 odd, 5 even. Use conditional probability.
<p>Odd: \(\{1,3,5,7,9,11\}\) -- 6 numbers. Even: \(\{2,4,6,8,10\}\) -- 5 numbers.</p><p>\(P(\text{sum even}) = \dfrac{\binom{6}{2}+\binom{5}{2}}{\binom{11}{2}} = \dfrac{15+10}{55} = \dfrac{25}{55} = \dfrac{5}{11}\).</p><p>\(P(\text{both odd}) = \dfrac{15}{55} = \dfrac{3}{11}\).</p><p>\(P(\text{both odd}|\text{sum even}) = \dfrac{3/11}{5/11} = \dfrac{3}{5}\)</p>
Correct Answer: C