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

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Manager
Joined: 14 Mar 2007
Posts: 233

Kudos [?]: 9 [0], given: 0

In a game with one die, player X wins if the number of [#permalink]

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24 Apr 2007, 05:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 9 [0], given: 0

Senior Manager
Joined: 11 Feb 2007
Posts: 350

Kudos [?]: 186 [0], given: 0

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24 Apr 2007, 06:27
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

Kudos [?]: 186 [0], given: 0

24 Apr 2007, 06:27
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# In a game with one die, player X wins if the number of

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