<p>A bag contains Red balls and Black balls such that \(4R + 6B = 10\). Two balls are drawn at random. If \(P\) is the probability that one ball is Red and one ball is Black, then \(P\) equals:</p>
Step-by-Step Solution
Key Concept: From the constraint 4R + 6B = 10, determine valid integer solutions for R and B, then calculate the hypergeometric probability of drawing exactly one red and one black ball from the total.
<p><strong>Step 1:</strong> Find valid values of R and B from 4R + 6B = 10.</p><p>Rearranging: 4R = 10 - 6B, so R = (10 - 6B)/4.</p><p>For R to be a non-negative integer: 10 - 6B ≥ 0 and 10 - 6B ≡ 0 (mod 4).</p><p>Testing values: B = 0 gives R = 2.5 (invalid); B = 1 gives R = 1 ✓; B ≥ 2 gives R < 0 (invalid).</p><p><strong>Step 2:</strong> With R = 1 and B = 1, total balls = 2.</p><p>P(one Red and one Black) = (C(1,1) × C(1,1))/C(2,2) = (1 × 1)/1 = 1.</p><p>∴ Answer: A (P = 1)</p>
Correct Answer: A