probability

Given a throw of three dice, let X,Y,Z be the number of dots showing on each. What is P(X<Y<Z)?

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  1. By symmetry this is

    1/3! P(X,Y,Z are all different)

    P(X,Y,Z are all different) =

    1- P(X,Y,Z are not all different)

    X,Y,Z are not all different if and only if one or more of the three conditions are satisfied:

    c1: X = Y

    c2: X = Z

    c3: Y = Z

    Terefore:

    P(X,Y,Z are not all different)=

    P(c1 OR c2 Or c3)

    The prinicple of Inclusion and Exclusion says that:

    P(c1 OR c2 Or c3) =

    P(c1) + P(c2) + P(c3) -

    [P(c1 And c2) + P(c1 And c3) +
    P(c2 And c3) ] +

    P(c1 And c2 And c3)

    We have P(c1) = P(c2) = P(c3) = 1/6

    and

    P(c1 And c2) = P(c1 And c3) =
    P(c2 And c3) = P(c1 And c2 And c3) = 1/36

    Therefore:

    P(c1 OR c2 Or c3) = 3/6 - 2/36 = 16/36 = 4/9

    P(X,Y,Z are all different) = 1-4/9 = 5/9

    P(X<Y<Z) = 1/6 * 5/9 = 5/54

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  2. Another way to compute this is by counting the number of ways you can make the three numbers diffrent. If you choose X first, then you have 6 choices for X. If Y is next then there are 5 choices for Y lft and 4 for Z. So, there are 6*5*4 possibilities. The probablity of any particular outcome is 1/6^3, so the probability is:

    6*5*4/6^3 = 20/36 = 5/9

    The probability that X, Y and Z are in the correct order is 1/6 times this probability which is:

    5/9 * 1/6 = 5/54

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