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Question

a, b, c throw a die alternatively by till one of them gets a 6 and wins a game.find the respective probabilities of winning if a starts first.
will we find probabilities of winning of all three i.e. a b and c .

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Solution

Let E be the event of getting a six in a single throw of the die.PE=16PE=1-16=56A wins if throws six in 1st or 4th or 7th or... throws. Also, B does not throw a six in 2nd or 5th or 8th... throws and C does not throw a six in 3rd or 6th or 9th... throws.PA wins=PEEEEEEEEEEEE...=PE+PEEEE+PEEEEEEE+...=PE+PEPEPEPE+PEPEPEPEPEPEPE+...=16+56316+56616+...=161-563=3691B wins if throws six in 2nd or 5th or 8th or... throws. Also, A does not throw a six in 1st or 4th or 7th... throws and C does not throw a six in 3rd or 6th or 9th... throws.PB wins=PEEEEEEEEEEEEEEE...=PEE+PEEEEE+PEEEEEEEE+...=PEPE+PEPEPEPEPE+PEPEPEPEPEPEPEPE+...=56×16+56416+56716+...=5361-563=3091Now,PA wins+ PB wins + PC wins=13691+3091+PC wins=1PC wins=1-3691-3091=2591

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