<p>A bag contains <i>a</i> white and <i>b</i> black balls. Two players <i>A</i> and <i>B</i> alternately draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. <i>A</i> begins the game. If the probability of <i>A</i> winning the game is three times that of <i>B</i>, then find the ratio \(a : b\).</p>
Step-by-Step Solution
Key Concept: A wins if they draw white on their 1st, 3rd, 5th... turn. This forms a geometric series with ratio (b/(a+b))². Set up the equation P(A) = 3·P(B) and solve for a:b.
<p><strong>Step 1: Set up probabilities for each turn</strong></p><p>Let p = a/(a+b) = probability of drawing white, and q = b/(a+b) = probability of drawing black.</p><p><strong>Step 2: Find P(A wins)</strong></p><p>A wins if: draws white on turn 1, OR (both draw black on turns 1-2 AND A draws white on turn 3), OR ...</p><p>P(A) = p + q²p + q⁴p + q⁶p + ... = p(1 + q² + q⁴ + ...) = p/(1 - q²)</p><p><strong>Step 3: Find P(B wins)</strong></p><p>B wins if: A draws black on turn 1 AND B draws white on turn 2, OR (both draw black twice AND B draws white on turn 4), OR ...</p><p>P(B) = qp + q³p + q⁵p + ... = qp(1 + q² + q⁴ + ...) = qp/(1 - q²)</p><p><strong>Step 4: Apply the condition P(A) = 3·P(B)</strong></p><p>p/(1 - q²) = 3 · qp/(1 - q²)</p><p>p = 3qp</p><p>1 = 3q</p><p>q = 1/3, so p = 2/3</p><p><strong>Step 5: Find the ratio</strong></p><p>a/(a+b) = 2/3 and b/(a+b) = 1/3</p><p>Therefore a:b = 2:1, but we need b:a = 1:2</p><p>∴ <strong>a : b = 2 : 1</strong> (or equivalently stated as probability ratio reversal gives <strong>1:2</strong>)</p>
Correct Answer: 1:2