Q. 27

Three persons A, B, C throw a die in succession till one gets a ‘six’ and wins the game. Find their respective probabilities of winning.

Answer :

Given that A,B and C throws a die.


The first who throw 6 wins the game.


P(S6) = P(getting 6)



P(SN) = P(not getting 6)




Let us assume A starts the game, A wins the game only when he gets 6 while throwing dice in 1st,4th,7th,…… times


Here the probability of getting sum 6 on throwing a dice is same for the players A, B and C


Since throwing a dice is an independent event, their probabilities multiply each other


P(Awins) = P(S9) + P(SN)P(SN)P(SN)P(S9) + P(SN)P(SN)P(SN)P(SN)P(SN)P(SN)P(S9) + ……………




The series in the brackets resembles the Infinite geometric series.


We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .






B wins the game only when he gets 6 while throwing dice in 2nd,5th,8th,…… times and others doesn’t get 6.


Since throwing a dice is an independent event, their probabilities multiply each other


P(Bwins) = (P(SN)P(S9)) + (P(SN)P(SN)P(SN)P(SN)P(S9)) + (P(SN)P(SN)P(SN)P(SN)P(SN)P(SN)P(SN)P(S9)) + ……………




The series in the brackets resembles the Infinite geometric series.


We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .






P(Cwins) = 1-P(Awins)-P(Bwins)




The probabilities of winning of A, B and C is .


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