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m05 #10

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Re: m05 #10 [#permalink] New post 10 Apr 2013, 09:54
I did it this way:

If sum = 8: first card could be 5 of the 6 (cards 2-6), second card must be 1 card depending on what the first was.

Probability of sum = 8: 5/6*1/6=5/36

If one card is 5 and sum is 8, the first could be 5, and the second must be 3, OR the first could be 3, and the second 5.

Probability 1 card is 5 and sum = 8: 1/6*1/6 + 1/6*1/6 = 2/36

So there are 5 possible instances where the sum is 8, 2 of which include a 5 i.e. D=2/5
Re: m05 #10   [#permalink] 10 Apr 2013, 09:54
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