Probability
Geometric Probability / Series
Grade 12
Question:
<p>A player \(X\) has a biased coin whose probability of showing heads is \(p\) and a player \(Y\) has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If \(X\) starts the game, and the probability of winning the game by both the players is equal, then the value of '\(p\)' is</p>
<p>\(\dfrac{1}{5}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{2}{5}\)</p>
<p>\(\dfrac{1}{4}\)</p>
Step-by-Step Solution
Key Concept: X wins on odd turns (1st, 3rd, 5th...) and Y wins on even turns (2nd, 4th, 6th...). Set up infinite geometric series for both players' winning probabilities and equate them since P(X wins) = P(Y wins).
<p><strong>Step 1:</strong> Identify winning scenarios</p><p>X wins if: Head on 1st turn OR (Tail on 1st, Tail on 2nd, Head on 3rd) OR (Tails × 4, Head on 5th)...</p><p>P(X wins) = p + (1-p)·(1/2)·p + [(1-p)·(1/2)]²·p + [(1-p)·(1/2)]³·p + ...</p><p><strong>Step 2:</strong> Sum the geometric series for X</p><p>P(X wins) = p[1 + (1-p)/2 + ((1-p)/2)² + ((1-p)/2)³ + ...]</p><p>P(X wins) = p · 1/(1 - (1-p)/2) = p/(1 - (1-p)/2) = p/((1+p)/2) = 2p/(1+p)</p><p><strong>Step 3:</strong> Find P(Y wins)</p><p>P(Y wins) = 1 - P(X wins) = 1 - 2p/(1+p) = (1+p-2p)/(1+p) = (1-p)/(1+p)</p><p><strong>Step 4:</strong> Equate probabilities</p><p>2p/(1+p) = (1-p)/(1+p)</p><p>2p = 1 - p</p><p>3p = 1</p><p>p = 1/3</p><p>∴ Answer: C (p = 1/3)</p>
Correct Answer: C