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Math Help - Please help on "get ball" issue, thanks

  1. #1
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    Please help on "get ball" issue, thanks

    5000 persons, 1500 balls, every person get 20 balls, what's the propobility that any one ball (among the 1500 balls) will be got by N persons? (N<=5000)
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    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by rabbituzi View Post
    5000 persons, 1500 balls, every person get 20 balls, what's the propobility that any one ball (among the 1500 balls) will be got by N persons? (N<=5000)
    You're going to have to explain this better. Does each person draw 20 balls, then put them all back for the next person to draw? Or is it that each time a person draws a ball they put it back before the next draw (i.e., its possible for a person to draw the same ball twice)?
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  3. #3
    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by rabbituzi View Post
    5000 persons, 1500 balls, every person get 20 balls, what's the propobility that any one ball (among the 1500 balls) will be got by N persons? (N<=5000)
    Assuming individuals pull 20 balls without replacment, then the probability that any one individual gets the magic ball is 20/1500. The probability that N people out of the 5000 get the magic ball is:

    <br />
p(N) = \left ( \frac {20} {1500} \right )^N \left ( \frac {1480} {1500} \right ) ^{ (5000-N) }  C_N ^ {5000-N}<br />
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    Thanks

    Thank you very much, ebaines. You help a lot!
    Last edited by mr fantastic; March 5th 2009 at 04:59 AM.
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    Maybe I have not explained my problem clearly,
    There are 5000 persons, 1500 balls.
    The ball is labeled with seriate number from 1 to 1500. And everyone selects 20 numbers randomly (not picking up balls but recording each number)
    What's the propobility that a ball with number X (X is a number between [1,1500]) will be selected by N persons? (N<=5000)

    I think the last part of ebaines's answer N!/(5000-N)! should be modified as
    5000!/N!

    If I am wrong, please help to correct , many thanks.
    Attached Thumbnails Attached Thumbnails Please help on &quot;get ball&quot; issue, thanks-.bmp  
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  6. #6
    MHF Contributor ebaines's Avatar
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    I am afraid that my original response was incorrect - apologies. The correct last term is:

    C(5000,N)

    Which can also be written as:

    <br />
\frac {5000!} {N!(5000-N)!}<br />

    Sorry for the confusion
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  7. #7
    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by rabbituzi View Post
    Maybe I have not explained my problem clearly,
    There are 5000 persons, 1500 balls.
    The ball is labeled with seriate number from 1 to 1500. And everyone selects 20 numbers randomly (not picking up balls but recording each number)
    What's the propobility that a ball with number X (X is a number between [1,1500]) will be selected by N persons? (N<=5000)

    I think the last part of ebaines's answer N!/(5000-N)! should be modified as
    5000!/N!

    If I am wrong, please help to correct , many thanks.
    An easy way to confirm that this isn't right is to test the case where N = 0 (that is, nobody picks the ball with a particular serial number). It's pretty clear that the probability of no one selecting a given ball is(1480/1500)^5000. But your formula gives (1480/1500)^5000 * 5000!
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  8. #8
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    You are correct , ebaines, I really appriciate your help.
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