Four persons A, B, C and D throw an unbiased die, turn by turn in succession till one gets an even number and win the game. What is the probability that $\mathrm{A}$ wins the game if A begins?

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Four persons A, B, C and D throw an unbiased die, turn by turn in succession till one gets an even number and win the game. What is the probability that $\mathrm{A}$ wins the game if A begins?

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kritika · Answer 1 · 2021-12-26T04:59:57+0000

Ans: (D)
Hint $: P($ even $)=\frac{1}{2}, P($ odd $)=\frac{1}{2}$
$P($ A wins $)=\frac{1}{2}+\left(\frac{1}{2}\right)^{4} \times \frac{1}{2}+\left(\frac{1}{2}\right)^{8} \times \frac{1}{2}+\left(\frac{1}{2}\right)^{12} \times \frac{1}{2}+\ldots \ldots \ldots \ldots \ldots \infty$
$$
=\frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^{4}}=\frac{1}{2} \times \frac{16}{15}=\frac{8}{15}
$$

Four persons A, B, C and D throw an unbiased die, turn by turn in succession till one gets an even number and win the game. What is the probability that \(\mathrm{A}\) wins the game if A begins?

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