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Germano, i need your help in one last PROBABILITY question. Thank you.?

Three numbers are selected at random (without replacement) from the first six positive integers. Let x donate the largest of the three numbers obtained. Find the probability distribution of X. Also, find the mean and variance of the distribution.

BA shall be yours.

Update:

The question is open for the house.

1 Answer

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  • ?
    Lv 7
    5 years ago
    Favorite Answer

    Well,

    okay, as the question is open "for the house" ... let's proceed

    X = Sup(X1, X2, X3)

    X1 has a uniform distribution within E = {1, 2, 3, 4, 5, 6}

    the mutual distribution of (X1, X2, X3) is, for any triplet of

    P( X1 = m, X2 = n, X3 = p) = 1/6 * 1/5 * 1/4 = 1/120

    by the way, for any two among X1, X2 and X3, for example X1 and X2 we have :

    P( X1 = m, X2 = n) = 1/6 * 1/5

    P(X = k) means ALL X1 and X2 and X3 are < k

    therefore :

    of course, we have :

    P(X= 1) = 0 and

    P(X= 2) = 0 as we draw three numbers without replacement

    P(X = 3) = P(X1 = 3, X2 = 1, X3 = 2) + P(X1 = 3, X2 = 2, X3 = 1) + ...

    ... + two terms for (X2 = 1) + ...

    ... + two terms for (X3 = 1) + ...

    = 6/120

    = 1/20

    P(X = 4) = P(X1 = 4, X2 = 1, X3 = 2) + P(X1 = 4, X2 = 2, X3 = 1) + ...

    ... P(X1 = 4, X2 = 1, X3 = 3) + P(X1 = 4, X2 = 3, X3 = 1) + ...

    ... P(X1 = 4, X2 = 2, X3 = 3) + P(X1 = 4, X2 = 3, X3 = 2) + ...

    ... + same 6 terms with (X2 =4) + ...

    ... + same 6 terms with (X3 =4) +

    = 3 * 3A2 * 1/120 <--- where 3A2 = 3! / (3 - 2)! = 6

    = 18/120

    = 3/20

    now,

    the number of terms (all equal) consists

    in choosing 1 among X1, X2 or X3 to represent the MAX and

    in arranging the remaining 2 among (4 - 1)

    so :

    P(X=5) = 3 * 4A2 * 1/120 <---- where 4A2 = 4! / (4 - 2)! = 12

    = 36/120

    = 6/20

    and finally :

    P(X=6) = 3* 5A2 * 1/120

    = 60/120

    = 10/20

    summary : we obtained the law :

    P(X= 1) = 0

    P(X= 2) = 0

    P(X=3) = 1/20

    P(X=4) = 3/20

    P(X=5) = 6/20

    P(X=6) = 10/20

    and we verify of course that the sum equals 1 !!

    the rest is just a matter of applying the formulae...

    E(X) = 3*P(X=3) + 4*P(X=4) + 5*P(X=5) + 6*P(X=6)

    = (1/20) * (3*1 + 4*3 + 5*6 + 6*10)

    = (1/20) * 105

    = 21/4 <------------ verify my calc !! I only guarantee the method !!

    Var(X) = E(X^2) - E(X)^2

    where :

    E(X^2) = 3^2*P(X=3) + 4^2*P(X=4) + 5^2*P(X=5) + 6^2*P(X=6)

    = (1/20) * (3^2*1 + 4^2*3 + 5^2*6 + 6^2*10)

    = 558/20

    and you can easily finish !!

    et voilà !! ;-)

    nice exercise ...

    hope it' ll help !!

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