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    sirpilf's Avatar
    sirpilf Posts: 1, Reputation: 1
    New Member

    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's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member

    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,
    reinsuranc's Avatar
    reinsuranc Posts: 92, Reputation: 6
    Junior Member

    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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