If A and B throw 2 dice such that if A gets a sum of 6, he wins and if B gets a sum of 7, he wins. Find the probability of there victory if A throws the dice first. They throw dices alternatively.
That might be simple...but plz try
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2 Answers
Prob of getting a sum 6 = 5/36
Prob of getting a sum 7= 6/36
Probability of not getting a sum 7 on the dice = 30/36=5/6
Hence the total probability of winning is given by the infinite series
5/36 + 31/36x5/6 x 5/36 + (31/36x5/6)2 x 5/36 + (31/36x5/6)3 x 5/36 + (31/36x5/6)4 x 5/36 .... .
= 5/36 {1 + 31/36x5/6 + (31/36x5/6)2 + (31/36x5/6)3 + (31/36x5/6)4 .... .}
= 5/36 {1 + 155/216 + (155/216)2 + (155/216)3 + (155/216)4 .... .}
= 5/361-155216 = 5/3661216 = 30/61