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

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Joined: 03 Feb 2003
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four fair dice are rolled. what is the probability of having [#permalink]  25 Jun 2003, 00:05
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four fair dice are rolled. what is the probability of having EXACTLY 1 three and EXACTLY 1 six?
Manager
Joined: 18 Jun 2003
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Is it not [25/(6^4)] x 16

= 1/6 x 1/6 x 5/6 x 5/6 x 4c1 x 4c1
Manager
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Posted: Wed Jun 25, 2003 5:49 pm Post subject:

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Is it not [25/(6^4)] x 16

= 1/6 x 1/6 x 5/6 x 5/6 x 4c1 x 4c1

I think the answer must be 1/6*1/6*4/6*4/6*4c1*4c1
You can not use three nor six, so the porbability for the other two dice are 4posibilities out of 6
SVP
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IMHO: 1/6*1/6*4/6*4/6*4C2=2/27

4/6 because 3 and 6 are both out
Manager
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stolyar wrote:
IMHO: 1/6*1/6*4/6*4/6*4C2=2/27

4/6 because 3 and 6 are both out

Could you explain why you multiply with 4C2 ?
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KSS wrote:
stolyar wrote:
IMHO: 1/6*1/6*4/6*4/6*4C2=2/27

4/6 because 3 and 6 are both out

Could you explain why you multiply with 4C2 ?

because EXACTLY 1 three and EXACTLY 1 six can go in different order, and we have count all of them.
Senior Manager
Joined: 11 Nov 2003
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I disagree with 2/27 as an answer. It should be 4/27.

Answer = 4C1 * (1/6) * 3C1 * (1/6) * (4/6)^2 = 4/27

The answer 2/27 is correct only if

3-6-1-2 is equal to 6-3-1-2. However, that should not be the case. They should be considered seperate outcomes.
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Gmatblast, I think you are right on this one. It took me a while to see the diff. between 4C1*3C1 and 4C2 but yes, the analysis is perfect. 4/27 should be the answer
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Best Regards,

Paul

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