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Math Help - a basket contains 5 red balls, 3 blue balls, 1 green balls

  1. #1
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    a basket contains 5 red balls, 3 blue balls, 1 green balls

    a basket contains 5 red balls, 3 blue balls, 1 green balls

    what is the probability of getting 2 red balls, 1 blue ball and 1 green ball if a sample of 4 balls are taken from the basket.

    How do i start solving this problem.

    Please help.
    Last edited by mr fantastic; May 27th 2010 at 08:00 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by ilikec View Post
    a basket contains 5 red balls, 3 blue balls, 1 green balls

    what is the probability of getting 2 red balls, 1 blue ball and 1 green ball if a sample of 4 balls are taken from the basket.

    How do i start solving this problem.

    Please help.
    Hi "i like c",

    there are 2 ways to work it out, if not more...

    If a sample of 4 balls are taken,
    we can list the order of selection if they are chosen one-by-one.

    RRBG red 1st, red 2nd, blue 3rd, green 4th.... P=\frac{5}{9}\ \frac{4}{8}\ \frac{3}{7}\ \frac{1}{6}
    RRGB
    RBRG
    RBGR
    RGRB
    RGBR
    GRRB
    GRBR
    GBRR
    BRRG
    BRGR
    BGRR

    If you calculate the remaining probabilities and sum them, you will have the
    probability of choosing 2 reds, 1 blue and 1 green.

    Alternatively there are \binom{9}{4} ways to choose 4 from 9.

    There are \binom{5}{2} ways to choose 2 reds

    There are \binom{3}{1} ways to choose one blue and only 1 way to choose the green.

    Any 2 reds can go with any blue, hence we multiply those numbers of ways
    to find the total number of ways we can get 2 reds, 1 blue and 1 green

    Hence you divide that number by the total number of ways to choose 4 from 9.
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  3. #3
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    How can you do it using Hyper-geometric Distribution principles?
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  4. #4
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    Using the multivariate hypergeometric distribution makes the problem much easier in my opinion.

    First let's label everything:

    N= total number of balls = 9
    n= total number of balls selected =4
    R = total number of red balls = 5
    B = total number of blue balls = 3
    G = total number of green balls =1
    r = total number of red balls selected = 2
    b = total number of blue balls selected = 1
    g = total number of green balls selected = 1

    Now simply solve:

    \frac{{R\choose r}{B\choose b}{G\choose g}}{{N\choose n}}

    As you can see, it's simply the combination of ways to pick 2 red balls out of 5 red balls multiplied by the combination of ways to pick 1 blue ball out of 3 blue ball multiplied by the combination of ways to select 1 green ball from 1 green ball all divide by the combination of ways to select 4 balls from 9 balls.
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