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Math Help - Probability question

  1. #1
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    Probability question

    I have 4 assignments to pass to three different people to do for me. What is the probability that each person will receive at least one assignment?

    Total combination: {4+3-1\choose 3-1} = {6\choose 2} = 15

    P(each person receive at least one assignment) = 3\over 15 = 1\over 5

    My answer is 0.2
    but it's wrong

    It's a MCQ question with options: (A) 8\over 9, (B) 64\over 81, (C) 4\over 9, (D) 16\over 81, (E) 5\over 9
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    I have 4 assignments to pass to three different people to do for me. What is the probability that each person will receive at least one assignment? It's a MCQ question with options: (A) 8\over 9, (B) 64\over 81, (C) \color{blue}4\over 9, (D) 16\over 81, (E) 5\over 9
    Consider that the 4 assignments are all distinct.
    There are 36 surjective (onto) functions from a set of 4 to a set of 3.
    There are 3^4 functions from a set of 4 to a set of 3.
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  3. #3
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    Thanks for the reply.

    Ya if the assignments are distinct, there will be a total of 3^4 possibilities.

    But how do you find the 36?

    My interpretation will be to permute 4 assignments among the 3 people (since each of them will receive one assignment), which is 4. 3. 2. 3 = 72, since the last assignment could be allocated to any one among the three. Therefore answer is A?
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  4. #4
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    Quote Originally Posted by noob mathematician View Post
    But how do you find the 36?
    The number of mappings from {1,2,3,4} to {a,b,c} which are onto is 36.
    Onto means that each of a, b, & c gets at least one.
    Look up the ways to count surjections.
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  5. #5
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    oic. Thanks a lot.

    The formula S(n,k)= \sum_{i=1}^{n} {k\choose i}<br />
(k-i)^n (-1)^i

    Ok!
    Last edited by noob mathematician; March 25th 2009 at 06:21 AM. Reason: Fixed latex
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  6. #6
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    Quote Originally Posted by noob mathematician View Post
    The formula S(n,k)= \sum_{i=1}^{n} {k\choose i}(k-i)^n (-1)^i
    That is close. But not exact.
    The number of surjections from a set of N elements to a set of K elements is:
    S(N,K) = \sum\limits_{j = 0}^K {\left( { - 1} \right)^j { K \choose j} \left( {K - j} \right)^N } .
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  7. #7
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    Oh ya.. haha some typo error there. Thanks for pointing out.
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