<p>If the probability of choosing an integer "\(k\)" out of \(2m\) integers \(1, 2, 3, \ldots, 2m\) is inversely proportional to \(k^4\) (\(1 \leq k \leq m\)). If \(x_1\) is the probability that chosen number is odd and \(x_2\) is the probability that chosen number is even, then</p>
Step-by-Step Solution
Key Concept: The probability distribution is split into two ranges with different functional forms: for k ≤ m, P(k) ∝ 1/k⁴, and for k > m (representing numbers m+1 to 2m), probabilities must be determined by the normalization constraint that all probabilities sum to 1.
<p><strong>Step 1:</strong> Set up the probability distribution. For 1 ≤ k ≤ m: P(k) = c/k⁴ where c is a constant.</p><p><strong>Step 2:</strong> For the remaining integers m+1, m+2, ..., 2m, let P(k) = d (constant probability for each).</p><p><strong>Step 3:</strong> Apply normalization: ∑P(k) = 1</p><p>$$\sum_{k=1}^{m} \frac{c}{k^4} + \sum_{k=m+1}^{2m} d = 1$$</p><p>$$c\sum_{k=1}^{m} \frac{1}{k^4} + md = 1$$</p><p><strong>Step 4:</strong> Calculate x₁ (probability of odd number):</p><p>For 1 ≤ k ≤ m: odd numbers are 1, 3, 5, ..., contribute c(1/1⁴ + 1/3⁴ + 1/5⁴ + ...)</p><p>For m+1 to 2m: odd numbers contribute d × (count of odd numbers in this range)</p><p><strong>Step 5:</strong> Calculate x₂ (probability of even number) similarly with even values.</p><p><strong>Step 6:</strong> The key insight is that for large m, the 1/k⁴ terms dominate for small k, and the constraint determines d. Since odd numbers are concentrated in the lower indices (where 1/k⁴ is larger), x₁ > x₂.</p><p>∴ Answer: A</p>
Correct Answer: A