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four fair dice are rolled. what is the probability of having

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four fair dice are rolled. what is the probability of having [#permalink] New post 24 Jun 2003, 23:26
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four fair dice are rolled. what is the probability of having exactly 2 threes?

(A) 1/36
(B) 1/72
(C) 1/6
(D) 1/3
(E) 25/216
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 [#permalink] New post 25 Jun 2003, 12:01
CAn you explain why you multiply by 4C2 and not 4?

thank you
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 [#permalink] New post 25 Jun 2003, 12:18
Don┬┤t explain, I have already figured it out
thank you
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 [#permalink] New post 25 Aug 2004, 08:52
Looking through some old problems and I was wondering if anyone could explain how 1/5*1/5*5/6*5/6*4C2=25/316 in this problem. I have no idea.
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 [#permalink] New post 25 Aug 2004, 09:12
can any one explain ?

probability of getting 3 and 3 from 2 dice = 1/36
the dice can be (1,2), (1,3), (1,4)(2,3)(2,4)3,4)
total prob . = 6/6^4
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 [#permalink] New post 25 Aug 2004, 12:16
the probability of 3 3 not3 not3 in that order is

1/6*1/6*5/6*5/6

now you want to consider all possible orderings that have 2 3's. Well, there are 4 dice, and you need 2 to be 3's, so you have 4C2 possible ways to roll 3's, times the probablity sequence above.
Joined: 31 Dec 1969
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 [#permalink] New post 25 Aug 2004, 12:22
Perhaps I missing the computation but how does that equal 25/216. I understand the logic but not the math.
Joined: 31 Dec 1969
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GMAT 1: 710 Q49 V38
GMAT 2: 660 Q V
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WE: Accounting (Accounting)
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 [#permalink] New post 25 Aug 2004, 12:27
Another stupid mistake on my part. 25/1296*6=150/1296 which simplifies to 25/216.
  [#permalink] 25 Aug 2004, 12:27
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