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# Four fair dice are rolled together. Find the probability of

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Four fair dice are rolled together. Find the probability of [#permalink]

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02 Jul 2004, 01:18
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Four fair dice are rolled together. Find the probability of having two 6s and two 1s.
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Re: PS: four fair dice [#permalink]

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02 Jul 2004, 06:34
stolyar wrote:
Four fair dice are rolled together. Find the probability of having two 6s and two 1s.

I'll try againg

P=fav/tot
fav=2C2*2C2
tot=6^4

P = 1/(9*36) = 1/324
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Re: PS: four fair dice [#permalink]

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02 Jul 2004, 06:41
stolyar wrote:
Four fair dice are rolled together. Find the probability of having two 6s and two 1s.

I'll try againg

P=fav/tot
fav=2C2*2C2
tot=6^4

P = 1/(9*36) = 1/324

P= 4C2 / (6^4)?

I think i got confused. Can you give the solution?
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02 Jul 2004, 06:42
2 6's and 2 4's can be chosen in 4C2 ways.

so the probability is 4C2 * (1/6)^2 * (1/6)^2 = 1/6 * 1/6 *1/6 = 1/216
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02 Jul 2004, 07:37
The other direct approach will be

Let A B C D be 4 dies

A B C D
6 6 1 1
6 1 6 1
.....
....
....
So on ie you have 4!/(2! *2!) of arranging them ie 6 ways.

Now Probaility = 1/6 * 1/6 * 1/6 * 1/6 * 6(ways) = 1/216
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02 Jul 2004, 07:47
sathya76 wrote:
The other direct approach will be

Let A B C D be 4 dies

A B C D
6 6 1 1
6 1 6 1
.....
....
....
So on ie you have 4!/(2! *2!) of arranging them ie 6 ways.

Now Probaility = 1/6 * 1/6 * 1/6 * 1/6 * 6(ways) = 1/216

Ok that is the solution i arrived in the end. Thanks for question.
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02 Jul 2004, 10:25
my approach:

1) a particular combination [6611]: 1/6*1/6*1/6*1/6
2) there are 4!/2!2!=6 of such combinations

P=1/(6*6*6)=1/216
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