Two persons A and B throw a die alternately till one of them gets a 'three' and wins the game. Find their respectively probabilities of winning, if a begins.

Explanation

Let E : person A gets three

F : person B gets three
`
P(E)=P(F)=frac{1}{6}, P(bar{E})=P(bar{F})=frac{5}{6}`
P( winning of A)`=frac{1}{6}+(frac{5}{6})^{2} frac{1}{6}+(frac{5}{6})^{4} frac{1}{6}+ldots`
`

=frac{frac{1}{6}}{1-(frac{5}{6})^{2}}=frac{6}{11}`


(sum of infinite terms of GP:` S=frac{a}{1-r} `)


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