<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: The precise key idea here is to recognize that for the sum of two numbers to be even, either both must be odd or both must be even. By calculating the probabilities of these events and applying the formula for conditional probability, we can find the desired 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>
<div class="key-concept"><strong>Key Concept:</strong> The precise key idea here is to recognize that for the sum of two numbers to be even, either both must be odd or both must be even. By calculating the probabilities of these events and applying the formula for conditional probability, we can find the desired probability.</div>
<div class="trap-box"><strong>Trap:</strong> Sum is even only when both numbers are of the same parity. $P(\text{sum even}) \neq \frac{1}{2}$ here because the set has unequal odd and even counts.</div>
Correct Answer: C