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Math Help - classic probability question...

  1. #1
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    classic probability question...

    8 balls are being put in 4 cells, randomly.

    1. What is the probability that all 8 balls will be put in 1 cell (doesn't matter which one) ?

    2. What is the probability that each cell had at least 1 ball

    3. What is the probability that in each cell there were exactly 2 balls ?

    4. What is the probability that all 8 balls were put in 2 cells only (doesn't matter which ones) ?

    Any help and guidance will be very appreciated !

    cheers !
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  2. #2
    MHF Contributor Unknown008's Avatar
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    The total number of ways the balls can arrange themselves in the 4 cells is 4^8

    Ball 1 has 4 possible ways, ball 2 has 4 possible ways, etc.

    1.

    Ball 1 can enter any cell. (4 ways)
    Ball 2 - 8 must go in that particular cell. (1 way each)

    2.

    This is 1 - P(one cell has no ball at all)

    Let's name the cells, cell 1, cell 2, cell 3 and cell 4.

    Ball 1 - 8 can enter any cell besides cell 1 (3^8 ways)
    Ball 1 - 8 can enter any cell besides cell 2 (3^8 ways)
    Ball 1 - 8 can enter any cell besides cell 3 (3^8 ways)
    Ball 1 - 8 can enter any cell besides cell 4 (3^8 ways)

    Total 4(3^8) ways

    Can you try those for now?

    EDIT: Had a later thought about 2.
    Last edited by Unknown008; January 16th 2011 at 09:24 AM.
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  3. #3
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    do you have the final solution to the qustion?
    Quote Originally Posted by WeeG View Post
    8 balls are being put in 4 cells, randomly.

    1. What is the probability that all 8 balls will be put in 1 cell (doesn't matter which one) ? (1/4)^8

    2. What is the probability that each cell had at least 1 ball? 1-P( one cell has no ball)=1-(3/4)^8

    3. What is the probability that in each cell there were exactly 2 balls ?

    4. What is the probability that all 8 balls were put in 2 cells only (doesn't matter which ones) ? (1/2)^8

    Any help and guidance will be very appreciated !

    cheers !
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  4. #4
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    I think the wording of question clearly implies that the balls are identical but the cells are distinct. If we read it in that way, then there are \dbinom{8+4-1}{8} ways to place the balls in the cells.

    There 4 ways to place all balls in one cell.

    There are \dbinom{4+4-1}{4} ways to place the balls in the cells so that each cell has at least one ball.

    Can you finish?
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  5. #5
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    Hi Guys, I solved 1 and 2, but stuck with 3 and 4.

    Maybe my question wasn't clear, I'll tell you how I solved 2, it will help you get the point.

    I said:

    A - Cell 1 is empty B - Cell 2 is empty C-Cell 3 is empty D-Cell 4 is empty

    Then I calculated the Union of events A,B,C and D (using the entire long formula of union with 4 events, with intersection and all that). Then the probability I was looking for is 1-the union. I did it, and got the correct answer.

    On 3 I need to get 0.038 and on 4 I need to get 0.023. I have no idea how....
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  6. #6
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    Are you counting the balls as if they are all distinct?
    Say the balls are numbered 1 to 8.

    What were the first two answers?
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  7. #7
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    I think they are numbered, does it matter ?

    The answer for "1" is 1/4^7 , since the first ball can go anywhere and then the rest has to go that 1 cell that was chosen.

    The answer for "2" is 0.62292, like I said, coming from the formula of union with 4 events reducing all intersections and adding some....(1-union)

    I am really lost about "3" and "4"...no clue
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  8. #8
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    I should have asked, "How do you know those answers are correct?"
    Are the answers given?


    I disagree with this way of counting these. But I know how they are gotten.
    The number of onto functions from a set of M to a set of N, M\ge N, is \text{Surj}(M,N) = \sum\limits_{k = 0}^N {( - 1)^k \binom{N}{k}\left( {N - k} \right)^M }.

    The answer to #4 is \dfrac{\binom{4}{2}\text{Surj}(8,2)}{4^8}.

    The answer to #3 is \dfrac{8!}{2^4\cdot 4^8}.
    Last edited by Plato; January 16th 2011 at 02:46 PM.
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  9. #9
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    yes, the answers are given.
    I am afraid I didn't understand the Surj thing....
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  10. #10
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    Thanks, I understand now what you did, and I got the correct answers, thank you !!
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