Two persons A and B throw a (fair)die (six-fa | vedclass

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Question

(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

Vedclass · Accepted Answer

$\frac{6}{7}$

Vedclass · Answer

(b)

Given, Probability of win $=\frac{5}{6}$

Probability of loss $=\frac{1}{6}$

Probability of $B$ wins =

$P(B)+P(\overline{B A} B)+P(\bar{B} \bar{A} \bar{B} \bar{A} B)+\ldots$

$=\frac{5}{6}+\frac{1}{6} 	imes \frac{1}{6} 	imes \frac{5}{6}+\frac{1}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} 	imes \frac{5}{6}+\ldots$

$=\frac{5}{6}\left(1+\frac{1}{6^2}+\frac{1}{6^4}+\frac{1}{6^6}+\ldotsight)$

$=\frac{5}{6}\left(\frac{1}{1-\frac{1}{36}}ight)=\frac{30}{35}=\frac{6}{7}$

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

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