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In a game with one die, player X wins if the number of

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In a game with one die, player X wins if the number of [#permalink] New post 24 Apr 2007, 04:19
In a game with one die, player X wins if the number of points on her throw is greater than, or equal to, the number of points on player YтАЩs throw. The probability of X winning is:


(A) ┬╜
(B) 2/3
(C) 7/12
(D) 13/24
(E) 19/36
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 [#permalink] New post 24 Apr 2007, 05:27
C.

Total outcomes 6*6 = 36
Cases in which X wins:
If Y=1, X=1~6
If Y=2, X=2~6
If Y=3, X=3~6
If Y=4, X=4~6
If Y=5, X=5,6
If Y=6, X=6
Total cases = 21

Therefore, 21/36 = 7/12

:)
  [#permalink] 24 Apr 2007, 05:27
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In a game with one die, player X wins if the number of

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