<p>An urn contains 3 red balls and \(n\) white balls. Mr. A draws two balls together from the urn. The probability that they have the same color is 1/2. Mr. B draws one ball from the urn, notes its color and replaces it. He then draws a second ball from the urn and finds that both balls have the same color is 5/8. The possible value of \(n\) is</p>
Step-by-Step Solution
Key Concept: Set up two probability equations based on drawing without replacement (Mr. A) and with replacement (Mr. B), then solve the resulting system to find n. The key is recognizing that 'same color' means both red OR both white in each case.
<p><strong>Step 1: Set up equation for Mr. A (without replacement)</strong></p><p>Mr. A draws 2 balls together. P(same color) = P(both red) + P(both white)</p><p>P(both red) = C(3,2)/C(3+n,2) = 3/[(3+n)(2+n)/2] = 6/[(3+n)(2+n)]</p><p>P(both white) = C(n,2)/C(3+n,2) = n(n-1)/2 / [(3+n)(2+n)/2] = n(n-1)/[(3+n)(2+n)]</p><p>Therefore: [6 + n(n-1)]/[(3+n)(2+n)] = 1/2</p><p>12 + 2n(n-1) = (3+n)(2+n)</p><p>12 + 2n² - 2n = 6 + 5n + n²</p><p>n² - 7n + 6 = 0</p><p>(n-1)(n-6) = 0 → n = 1 or n = 6</p><p><strong>Step 2: Set up equation for Mr. B (with replacement)</strong></p><p>With replacement: P(same color) = P(both red) + P(both white)</p><p>P(both red) = (3/(3+n))²</p><p>P(both white) = (n/(3+n))²</p><p>Therefore: (3/(3+n))² + (n/(3+n))² = 5/8</p><p>(9 + n²)/(3+n)² = 5/8</p><p>8(9 + n²) = 5(3+n)²</p><p>72 + 8n² = 5(9 + 6n + n²)</p><p>72 + 8n² = 45 + 30n + 5n²</p><p>3n² - 30n + 27 = 0</p><p>n² - 10n + 9 = 0</p><p>(n-1)(n-9) = 0 → n = 1 or n = 9</p><p><strong>Step 3: Find common solution</strong></p><p>From Mr. A: n = 1 or 6</p><p>From Mr. B: n = 1 or 9</p><p>Common value: n = 1</p><p>∴ The possible value of n is <strong>1</strong></p>
Correct Answer: D