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-   -   Probability of three fair dice (https://www.askmehelpdesk.com/showthread.php?t=318734)

  • Feb 17, 2009, 06:59 PM
    sirpilf
    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?
  • Feb 18, 2009, 06:10 AM
    galactus
    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,
  • Feb 20, 2009, 04:15 PM
    reinsuranc
    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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