Probability mass funcitons

**Andreamet** · March 26th 2008, 03:41 PM

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

**mr fantastic** · March 27th 2008, 02:45 AM

Originally Posted by Andreamet

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.

**mr fantastic** · March 27th 2008, 02:58 AM

Originally Posted by mr fantastic

[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,

$\Pr(R = r) = {12 \choose r} \left( \frac{4}{19}\right)^r \left( \frac{15}{19}\right)^{12-r}$ .

**mr fantastic** · March 31st 2008, 05:35 AM

Originally Posted by Andreamet

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

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