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    niko's Avatar
    niko Posts: 10, Reputation: 1
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    #1

    Aug 13, 2005, 12:15 PM
    Probability
    I have another good probability question that I can't answer.

    An ordinary six-sided die is rolled 12 times. If X1 is the number of 1's and X2 is the number of 2's and so on, give the probability for:

    a. P(X1=2, X2=3, X3=1, X4=0, X5=4, X6=2)

    b. P(X1=X2=X3=X4=X5=X6)
    niko's Avatar
    niko Posts: 10, Reputation: 1
    New Member
    		  
    #2

    Aug 13, 2005, 01:27 PM
    Never mind, I figured it out.

    The answer is

    12!/(2!3!4!2!) * (1/6)^12


    Thanks anyway!
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