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  • Need explanation for: 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 playalternately. The player who throws a head first is a winner. If X starts

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 :

  • Option 1)

    \frac{1}{5}

  • Option 2)

    \frac{1}{3}

  • Option 3)

    \frac{2}{5}

  • Option 4)

    \frac{1}{4}

 

Answers (1)

best_answer

As we learned,

@2981

X wins when we have

H, TTH, 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{\left (1-P \right )}{3}

Since P(x)=P(y) we get

\frac{4P}{3}=\frac{2\left ( 1-P \right )}{3}\: \Rightarrow \: P=\frac{1}{3}


Option 1)

\frac{1}{5}

Option 2)

\frac{1}{3}

Option 3)

\frac{2}{5}

Option 4)

\frac{1}{4}

Posted by

perimeter

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