# Probability mass funcitons

• March 26th 2008, 03:41 PM
Andreamet
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
• March 27th 2008, 02:45 AM
mr fantastic
Quote:

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.
• March 27th 2008, 02:58 AM
mr fantastic
Quote:

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.

(Rofl) Well, I suppose to ease my conscience I should point out that this marginal pmf is binomial (Rofl) .....

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,

$\Pr(R = r) = {12 \choose r} \left( \frac{4}{19}\right)^r \left( \frac{15}{19}\right)^{12-r}$.
• March 31st 2008, 05:35 AM
mr fantastic
Quote:

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