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 sirpilf Posts: 1, Reputation: 1 New Member #1 Feb 17, 2009, 06:59 PM
probability of three fair dice
I am taking stats modeling, its suppose to be a 5000 level course, but right now I have doing review homework, its been 2 years since prob, and this problem is driving me crazy. It should only take 2 min and I can't for gods sake get it.

roll three 6 sided fair dice, what's probability that you get exactly 2 dice with the same number, so the 3rd dice must be a different number.

in this first chapter, we have not done binomials yet, and I thought it was just 1 * (1/6) * (5/6) = 5/36. But its wrong, how do I do this?
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Feb 18, 2009, 06:10 AM
Don't forget your factorial. There are to the same out of three. It could be the first and second, second and third or first and third. Multiply by 3.

We have 3 die and 2 are the same, $\frac{3!}{2!}=3$
 reinsuranc Posts: 92, Reputation: 6 Junior Member #3 Feb 20, 2009, 04:15 PM
The probability of getting exactly two 1's is 3*[(1/6)^2]*(5/6) = 15/216.

(The 3 takes care of which of the three dice gives doubles - there are 3 possibilities.

The probability of getting exactly two of any number is 6 times that, or 90/216.

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