Two persons $A$ and $B$ are throwing an unbiased six faced dice alternatively, with the condition that the…
Two persons $A$ and $B$ are throwing an unbiased six faced dice alternatively, with the condition that the person who throws 3 first wins the game. If $A$ starts the game, then probabilities of $A$ and $B$ to win the same are, respectively
$\frac{6}{11}, \frac{5}{11}$
$\frac{5}{11}, \frac{6}{11}$
$\frac{8}{11}, \frac{3}{11}$
$\frac{3}{11}, \frac{8}{11}$
Solution
Here, probability of success,
$p=\frac{1}{6}$ and probability of failure. $q=\frac{5}{6}$
Let $A$ starts the game, then
$\begin{aligned}
& A=\frac{1}{6}+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^4+\ldots . . \\
& P(A)=\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{\frac{1}{6}}{1-\frac{25}{36}} \\
& =\frac{\frac{1}{6}}{\frac{11}{36}}=\frac{6}{11}
\end{aligned}$
Total probability, $P(A)+P(B)=1$
$\begin{aligned}
& \therefore \quad P(B)=1-P(A) \\
& =1-\frac{6}{11}=\frac{5}{11}
\end{aligned}$