Probability
Conditional Probability
Grade 12
Question:
<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>
<p>(a) \(\frac{2}{5}\)</p>
<p>(b) \(\frac{1}{2}\)</p>
<p>(c) \(\frac{7}{10}\)</p>
<p>(d) \(\frac{3}{5}\)</p>
Step-by-Step Solution
Key Concept: For an even sum, both numbers must have the same parity. Count favorable outcomes (both even) and condition on the sample space (even sum).
<p><strong>Step 1:</strong> For the sum to be even, either both numbers are even or both are odd.</p><p><strong>Step 2:</strong> From $\{1, 2, \ldots, 11\}$, there are 5 even numbers $\{2, 4, 6, 8, 10\}$ and 6 odd numbers $\{1, 3, 5, 7, 9, 11\}$.</p><p><strong>Step 3:</strong> Number of ways to select both even: $\binom{5}{2} = 10$.</p><p><strong>Step 4:</strong> Number of ways to select both odd: $\binom{6}{2} = 15$.</p><p><strong>Step 5:</strong> Total ways with even sum: $10 + 15 = 25$.</p><p><strong>Step 6:</strong> $P(\text{both even | sum is even}) = \frac{10}{25} = \frac{2}{5}$.</p>
Correct Answer: A