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Math Help - Simple probability question

  1. #1
    MHF Contributor alexmahone's Avatar
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    Simple probability question

    If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than one ball?

    Please give only a hint, and not the full solution.
    Thanks from mash
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  2. #2
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    Re: Simple probability question

    Quote Originally Posted by alexmahone View Post
    If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than one ball?Please give only a hint, and not the full solution.
    If we know that the boxes are distinct then there are \binom{12+20-1}{12} ways to put 12 identical balls into twenty different boxes.
    That is a start for you.

    Now if there is a different reading of this question, let us know.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Simple probability question

    Quote Originally Posted by Plato View Post
    If we know that the boxes are distinct then there are \binom{12+20-1}{12} ways to put 12 identical balls into twenty different boxes.
    That is a start for you.
    Thanks, but how did you get \binom{12+20-1}{12}?
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    Re: Simple probability question

    Quote Originally Posted by alexmahone View Post
    Thanks, but how did you get \binom{12+20-1}{12}?
    That is a standard occupancy formula.
    The number of ways to put N identical objects into K distinct cells is
    \binom{N+K-1}{N}.

    BTW. Why are you asked to do problems that you seem totally unprepared to attempt?
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Re: Simple probability question

    Quote Originally Posted by Plato View Post
    BTW. Why are you asked to do problems that you seem totally unprepared to attempt?
    I'm really sorry but I suck at combinatorics. I'm currently teaching myself Probability and Statistics from DeGroot and got stuck at this problem.
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Re: Simple probability question

    Quote Originally Posted by Plato View Post
    If we know that the boxes are distinct then there are \binom{12+20-1}{12} ways to put 12 identical balls into twenty different boxes.
    The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, isn't the total number of ways to put 12 balls into 20 boxes 20^{12}?
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    Re: Simple probability question

    Quote Originally Posted by alexmahone View Post
    The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, isn't the total number of ways to put 12 balls into 20 boxes 20^{12}?
    Are you assuming that the balls are all different?
    That does not seem to be implied in the statement of the question.
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Re: Simple probability question

    Quote Originally Posted by Plato View Post
    Are you assuming that the balls are all different?
    That does not seem to be implied in the statement of the question.
    How does that make a difference, when all we're asked is to find the probability that no box will receive more than one ball?
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  9. #9
    MHF Contributor alexmahone's Avatar
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    Re: Simple probability question

    My solution:

    No. of favourable arrangements:
    The 1st ball can be put into any of the 20 boxes, the 2nd ball can can be put into any of the other 19 boxes and so on. So, the number of ways to put 12 balls into 20 boxes so that no box receives more than one ball is 20*19*\cdots*9=\frac{20!}{8!}.

    Total no. of arrangements:
    The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, the total number of ways to put 12 balls into 20 boxes is 20^{12}.

    So, P=\frac{20!}{8!20^{12}}
    Last edited by alexmahone; March 30th 2012 at 05:18 PM.
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    Re: Simple probability question

    Quote Originally Posted by alexmahone View Post
    My solution:

    No. of favourable arrangements:
    The 1st ball can be put into any of the 20 boxes, the 2nd ball can can be put into any of the other 19 boxes and so on. So, the number of ways to put 12 balls into 20 boxes so that no box receives more than one ball is 20*19*\cdots*9=\frac{12!}{8!}.

    Total no. of arrangements:
    The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, the total number of ways to put 12 balls into 20 boxes is 20^{12}.

    So, P=\frac{12!}{8!20^{12}}
    I think you have a great deal to learn about modeling counting problems.
    You have answered this question: If we pick a function from a set of ten to a set of twenty, what is the probability that function is an injection?

    Does that model what this question is asking?
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  11. #11
    MHF Contributor alexmahone's Avatar
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    Re: Simple probability question

    Quote Originally Posted by Plato View Post
    You have answered this question: If we pick a function from a set of twelve to a set of twenty, what is the probability that function is an injection?

    Does that model what this question is asking?
    Yes.

    BTW, my answer matches with the one given at the back of the textbook. Why do you think it is wrong?
    Last edited by alexmahone; March 30th 2012 at 04:12 PM.
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  12. #12
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    Re: Simple probability question

    Quote Originally Posted by alexmahone View Post
    Yes.
    BTW, my answer matches with the one given at the back of the textbook. Why do you think it is wrong?
    I don't think that the answer is wrong. I think the model is wrong.
    Why not assume that the boxes are identical and the balls are different?
    OR, Why not assume that the boxes are identical and the balls are identical?

    These are legitimate questions asked in counting theory.

    How does one model these sorts of questions.
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