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Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is
$\frac{5}{6}$
$\frac{6}{7}$
$\frac{7}{8}$
$\frac{8}{9}$
Solution
(b)
Given, Probability of win $=\frac{5}{6}$
Probability of loss $=\frac{1}{6}$
Probability of $B$ wins =
$P(B)+P(\overline{B A} B)+P(\bar{B} \bar{A} \bar{B} \bar{A} B)+\ldots$
$=\frac{5}{6}+\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}+\ldots$
$=\frac{5}{6}\left(1+\frac{1}{6^2}+\frac{1}{6^4}+\frac{1}{6^6}+\ldots\right)$
$=\frac{5}{6}\left(\frac{1}{1-\frac{1}{36}}\right)=\frac{30}{35}=\frac{6}{7}$
