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Math Help - Probability mass funcitons

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    Probability mass funcitons

    a jury of 12 people is randomly selected from a group of 8 red hat, 7 blue hat,3 green hat and 20 yellow hat potential jurors. Let R, B, G and Y be the # of jurors, respectively. Calculate the joit probability mass function of R,B,G,Y and the marginal probability mass function of R
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    Quote Originally Posted by Andreamet View Post
    a jury of 12 people is randomly selected from a group of 8 red hat, 7 blue hat,3 green hat and 20 yellow hat potential jurors. Let R, B, G and Y be the # of jurors, respectively. Calculate the joit probability mass function of R,B,G,Y and the marginal probability mass function of R
    You're dealing with a multinomial distribution with n = 12, p_R = \frac{8}{38} = \frac{4}{19}, p_B = \frac{7}{38}, p_G = \frac{3}{38} and p_Y = \frac{20}{38} = \frac{10}{19}.

    The joint pmf for R, B, G and Y is therefore given by


    p(r, b, g, y) = \frac{12!}{r!\, b! \, g! \, y!} \, \left( \frac{4}{19} \right)^r \, \left( \frac{7}{38} \right)^b \, \left( \frac{3}{38} \right)^g \, \left( \frac{10}{19} \right)^y


    where r + b + g + y = 12.


    The marginal pmf of R is found by calculating

    \Pr(R = r) = \sum_b \sum_g \sum_y p(r, b, g, y)


     = \sum_b \sum_g \sum_y \frac{12!}{r!\, b! \, g! \, y!} \, \left( \frac{4}{19} \right)^r \, \left( \frac{7}{38} \right)^b \, \left( \frac{3}{38} \right)^g \, \left( \frac{10}{19} \right)^y


    where b + g + y = 12 - r.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    [snip]
    The marginal pmf of R is found by calculating

    \Pr(R = r) = \sum_b \sum_g \sum_y p(r, b, g, y)


     = \sum_b \sum_g \sum_y \frac{12!}{r!\, b! \, g! \, y!} \, \left( \frac{4}{19} \right)^r \, \left( \frac{7}{38} \right)^b \, \left( \frac{3}{38} \right)^g \, \left( \frac{10}{19} \right)^y


    where b + g + y = 12 - r.
    Well, I suppose to ease my conscience I should point out that this marginal pmf is binomial .....

    As you select hats you'll have successes and failures with selecting red ones. It's not hard to see then that

    R ~ Binomial \left( n = 12, \, p = \frac{4}{19} \right), that is,

    \Pr(R = r) = {12 \choose r} \left( \frac{4}{19}\right)^r \left( \frac{15}{19}\right)^{12-r}.
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  4. #4
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    Quote Originally Posted by Andreamet View Post
    a jury of 12 people is randomly selected from a group of 8 red hat, 7 blue hat,3 green hat and 20 yellow hat potential jurors. Let R, B, G and Y be the # of jurors, respectively. Calculate the joit probability mass function of R,B,G,Y and the marginal probability mass function of R
    Since the jurors are chosen (I assume) without replacement, the distribution will actually be the multivariate hypergeometric distribution:

    p(r, b, g, y) = \frac{ {8 \choose r} {7 \choose b} {3 \choose g} {20 \choose y}}{{ 38 \choose 12}}

    where r + b + g + y = 12.

    You could also build in the r + b + g + y = 12 restriction and write the pmf as:

    p(r, b, g, y) = \frac{ {8 \choose r} {7 \choose b} {3 \choose g} {20 \choose {12 - r - b - g} }}{{ 38 \choose 12}}.


    The marginal pmf for red will be a hypergeometric distribution:

    \Pr (R = r) = \frac{ {8 \choose r} {30 \choose {12 - r}} }{{38 \choose 12}}.


    Sorry for the original mistake in interpretation
    Last edited by mr fantastic; March 31st 2008 at 05:54 AM.
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