<p>Consider a game played by 10 people in which each flips a fair coin at the same time. If all but one of the coins comes up the same, then the odd person wins (e.g., if there are nine tails and one head then person having head wins). If such a situation does not occur, the players flip again. Find the probability that game is settled on or after <em>n</em>th toss.</p>
Step-by-Step Solution
Key Concept: The game settles when exactly one person differs from the other nine. Calculate P(settle in one round) = P(9 heads, 1 tail) + P(9 tails, 1 head) = 10·(1/2)^10 + 10·(1/2)^10 = 5/128, then use geometric distribution for settling on or after nth toss.
<p><strong>Step 1:</strong> Find probability that game settles in a single round (one person different from other 9).</p><p>Total ways to get 9 heads and 1 tail: C(10,1) = 10</p><p>Total ways to get 9 tails and 1 head: C(10,1) = 10</p><p>Total possible outcomes: 2^10</p><p>P(settle) = (10 + 10)/2^10 = 20/1024 = 5/256</p><p><strong>Step 2:</strong> Verify: 5/256 = 5/2^8 ✓</p><p><strong>Step 3:</strong> For the game to settle on or after the nth toss:</p><p>• Game must NOT settle in tosses 1, 2, ..., (n-1): probability = (1 - 5/2^8)^(n-1)</p><p>• Game MUST settle on toss n: probability = 5/2^8</p><p><strong>Step 4:</strong> Apply multiplication rule (independent events):</p><p>P(settle on or after nth toss) = (1 - 5/2^8)^(n-1) · 5/2^8</p><p><strong>Note:</strong> This is the standard geometric distribution formula P(X ≥ n) = (1-p)^(n-1)·p where p = 5/2^8 is the success probability.</p>
Correct Answer: \(\left(1 - \dfrac{5}{2^8}\right)^{n-1} \cdot \dfrac{5}{2^8}\)