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Math Help - permutation problem

  1. #1
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    Apr 2005
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    11

    permutation problem

    The problem is from DeGroot 3rd edition, section 1.7,
    problem 10: n=100 balls, r red balls. The balls
    are chosen one at a time without replacement.
    a) what is the pr that a red ball is chosen on the
    first selection? b) what is the pr a red ball is
    chosen on the 50th selection? c) what is the pr
    a red ball is chosen on the 100th selection?

    a) Pr(red ball 1st) = r/n, easy

    c) Pr(red ball 100th)? There are 100!
    arrangements of the n=100 balls because sampling
    is w/o replacement. Of those arrangements, the
    only way to guarantee a red ball is chosen last
    is to arrange the rest of the r-1 red balls in
    the first 99 positions, which may be done in
    P(99,r-1) ways where P(n,k) is the permutation
    function. Therefore, Pr(red ball 100th) =
    P(99,r-1) / 100!

    b) Pr(red ball 50th)? Not so sure of this one.
    I don't think I can use the above reasoning
    because r might be > 50. There are a total of
    100! permutations in 100 selections, 50! by the
    50th selection. If r<=50 then there are P(50,r)
    permutations of the red balls. To guarantee the
    50th is a red ball, there are P(49,r-1) ways to
    arrange the red balls. If r=50, then Pr(50th
    red)=1. If r>50, then Pr(50th red)=1. For r<50
    Pr(50th red) should be P(49,r-1)/50! But how to
    combine the three cases for r to get one pr for
    all r? I'm fairly sure I'm missing something
    here.
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  2. #2
    Newbie
    Joined
    Oct 2005
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    Smile Red Ball Selection

    You might want to consider using a PR value from a
    Hypergeometric Distribution.
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  3. #3
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    Joined
    Apr 2005
    Posts
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    I feel really dumb posting the solution becauses of all the hoops I jumped through. My professor said that the Pr(Red 1st) = Pr(Red 50th) = Pr(Red 100th) = r/100. Given no other information, the probability of a red ball on each selection is the same. Even sampling without replacement. Go figure. Trick question!
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