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JEE Mains · Maths · STD 12 - 13. probability
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
- A \(\frac{1}{3}\)
- B \(\frac{1}{5}\)
- C \(\frac{1}{4}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(X\) wins when we have \(\mathrm{H}, \mathrm{TTH}, \mathrm{TTTTH}\) So probability that \(X\) wins \(=P+\frac{P}{4}+\frac{P}{16} \cdots \cdots=4 \frac{P}{3}\) \(Y\) wins \(=\frac{1-P}{2}+\frac{1-P}{8}+\frac{1-P}{32}=2 \frac{(1-P)}{3}\) Since \(P(x)=P(y)\) we get…
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