Probability
Coin Tossing
Grade 12
Question:
<p>In a game a coin is tossed \(2n + m\) times and a player wins if he does not get any two consecutive outcomes same for at least \(2n\) times in a row. The probability that player wins the game is</p>
<p>(1) \(\frac{m+2}{2^{2n}+1}\)</p>
<p>(2) \(\frac{2n+2}{2^{2n}}\)</p>
<p>(3) \(\frac{2n+2}{2^{2n+1}}\)</p>
<p>(4) \(\frac{m+2}{2^{2n}}\)</p>
Step-by-Step Solution
Key Concept: Model consecutive outcomes as an alternating sequence problem. A player wins if there exist at least 2n consecutive tosses where no two adjacent outcomes are identical (alternating pattern H-T-H-T... or T-H-T-H...). Use recurrence relations or state-based probability to count favorable sequences.
<p><strong>Step 1:</strong> Define the winning condition precisely. Player wins if there exists a substring of 2n consecutive tosses with alternating outcomes (no two consecutive same). Total tosses = 2n + m.</p><p><strong>Step 2:</strong> For any substring of length 2n to alternate, there are exactly 2 favorable patterns: HTHTH... and THTH... (each has probability (1/2)^(2n)).</p><p><strong>Step 3:</strong> The total number of possible outcomes for (2n+m) tosses is 2^(2n+m).</p><p><strong>Step 4:</strong> Count sequences with at least one alternating block of length 2n. There are (m+1) possible positions where a block of length 2n can start (positions 1, 2, ..., m+1). By inclusion-exclusion or direct counting, favorable sequences = 2·(m+1)·2^m (accounting for the 2 alternating patterns and arrangements of remaining m tosses).</p><p><strong>Step 5:</strong> Probability = [2(m+1)·2^m] / 2^(2n+m) = 2(m+1)·2^m / 2^(2n+m) = (m+1)·2^(m+1) / 2^(2n+m) = (m+1) / 2^(2n-1).</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C