Probability

[匿名]

2010-12-24 6:48 pm

May and Amy throw an 8 - sided dice by turn with May throws first. Who get the '8' first will win. Given that Amy wins the game, find the probability that she did not win the game before the 2nd throw.

更新1:

To :自由自在 Description of the last 2 lines are the same but the answers are not the same?? Please clarify, thanks. Is it May and Amy mixed up?

回答 (1)

用戶9112

2010-12-24 7:04 pm

✔ 最佳答案

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2010-12-24 11:40:32 補充：
The last line should be revised as "Conditional probability that she does not win before the 2nd throw given she wins"

2010-12-24 12:05:36 補充：
Yes, I mixed up Amy with May.
Amy wins at the first throw = (7/8)(1/8) as May does not win at first throw
Amy wins at the 2nd throw = (7/8)(7/8)(7/8)(1/8)
Amy wins at the 3rd throw = [(7/8)(7/8)]^2 (7/8)(1/8)
....
Hence probability that Amy wins = 7/64[1/(1-49/64)]=7/15

2010-12-24 12:05:42 補充：
Conditional probability that Amy wins before 2nd throw given she wins = (7/64) / (7/15) = Conditional probability that Amy does not win before the 2nd throw given she wins
= 1 - 15/64 = 49/64

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