Q78 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

Two players A and B are alternately throwing a coin and a die together. A player who first throws head and 6 wins the game. If A starts the game, then the probability that B wins the game is.

  • A
    $\frac{12}{23}$
  • Correct answer
    B
    $\frac{11}{23}$
  • C
    $\frac{5}{119}$
  • D
    $\frac{12}{119}$

Solution

Let $P(B)$ to wins the game is $=\frac{1}{12}$ Since, B can win the game in 2nd throw, 4th throw, 6th throw and so on. Hence $P(B)$ to win the game is $$ \begin{aligned} & =\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})+\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})+\ldots \ldots \infty \\ & =\frac{11}{12} \times \frac{1}{12}+\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12}+\ldots \ldots \ldots . \\ & =\frac{1}{12}\left[\frac{11}{12}+\left(\frac{11}{12}\right)^3+\left(\frac{11}{12}\right)^5+\ldots \ldots .\right] \\ & =\frac{1}{12} \times \frac{\frac{11}{12}}{1-\left(\frac{11}{12}\right)^2}=\frac{11}{23} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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