<p>A biased coin with probability <i>p</i>, 0 < <i>p</i> < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2/5, then <i>p</i> equals</p>
Step-by-Step Solution
Key Concept: For the first head to appear on an even toss, we need (T,H), (T,T,T,H), (T,T,T,T,T,H), etc. This forms a geometric series: (1-p)p + (1-p)³p + (1-p)⁵p + ... = p(1-p)[1 + (1-p)² + (1-p)⁴ + ...], which sums to p(1-p)/(1-(1-p)²).
<p><strong>Step 1:</strong> Let q = 1-p (probability of tails). First head on even toss means: (T,H), (T,T,T,H), (T,T,T,T,T,H), ...</p><p><strong>Step 2:</strong> P(even number of tosses) = qp + q³p + q⁵p + ... = p·q(1 + q² + q⁴ + ...) = pq/(1-q²)</p><p><strong>Step 3:</strong> Since 1-q² = 1-(1-p)² = 2p-p² = p(2-p), we have: P(even) = pq/[p(2-p)] = q/(2-p) = (1-p)/(2-p)</p><p><strong>Step 4:</strong> Set equal to 2/5: (1-p)/(2-p) = 2/5</p><p><strong>Step 5:</strong> Cross multiply: 5(1-p) = 2(2-p) → 5-5p = 4-2p → 1 = 3p → p = 1/3</p><p><strong>Verification:</strong> (1-1/3)/(2-1/3) = (2/3)/(5/3) = 2/5 ✓</p><p>∴ Answer: A (p = 1/3)</p>
Correct Answer: A