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  • Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that <img alt="\mathrm{A}" src="https:/

Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that \mathrm{A} wins the game is

Option: 1

\frac{5}{11}


Option: 2

\frac{1}{2}


Option: 3

\frac{7}{13}


Option: 4

\frac{6}{11}


Answers (1)

best_answer

\mathrm{P(6) =1 / 6}
\mathrm{P(1 / 2 / 3 / 4 / 5) =5 / 6}

The case can be
\mathrm{A\: or\: \bar{A} \bar{B} A\: or \: \bar{A} \bar{B} \bar{A} \bar{B} A\: \&\: so \: on.}

where        \mathrm{A =1 / 6}
                \mathrm{\bar{A} =\bar{B}=5 / 6}

\mathrm{\therefore P=\frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\cdots-\infty}
         \mathrm{=\frac{1}{6}\left(1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\cdots-+\infty\right) }
          \mathrm{=\frac{1 / 6}{1-\left(\frac{5}{6}\right)^{2}}=\frac{1 / 6}{1-25 / 36} }
         \mathrm{=\frac{1 / 6}{11 / 36}=\frac{6}{11}}
         

Posted by

Pankaj Sanodiya

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