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Two person A and B toss an unbiased coin alternatelyon the understanding that the first who gets the headwins. If A starts the game, find their respective chanceof winning. |
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Answer» 'A' can win in odd trials. Let the probability of A getting heads be 'p' So '1-p' is the probability that 'A' loses. If 'A' wins in 1st trail then it is simply = p for 'A' to win in 3rd it has to lose in 1st two and win in 3rd = (1-p)^2 *p..and so on. So prob of A winning = p + (1-p)^2*p + (1-p)^4*p+... = p [1+(1-p)^2 + (1-p)^4+... ] p=1/2 So, probability of 'A winning' = 1/2 * [4/(4-1)] [summing the geometric series] = 2/3 Thus, probability of 'B' winning is 1/3 |
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