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1. A and B throw alternatively with a pair of dice. A wins if he throws 6 before 8 throws 7 andB wins if he throws seven before A throws 6. A begins the game. Show that odds in avourof A are 30 : 31. |
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Answer» (A)30/61 The reason being: The probability of A throwing ‘6’ is 5/36. So, the probability of A not throwing ‘6’ is 1 – 5/36 = 31/36. The probability of B throwing ‘7’ is 6/36. So, the probability of B not throwing ‘7’ is 1 – 6/36 = 30/36. The probability that somebody wins is [1-{(31/36)(30/36)}]. So, The probability of winning of A provided that somebody wins = (5/36)/[1-{(31/36)(30/36)}] = 30/61. |
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