Joint vs marginal distribution of black and white balls from an urn

[michaelxavier]

Homework Statement

An urn contains $p$ black balls, $q$ white balls, and $r$ red balls; and $n$ balls are chosen without replacement.
a. Find the joint distribution of the numbers of black, red, and white balls in the sample.
b. Find the joint distribution of the numbers of black and white balls in the sample.

Homework Equations

The Attempt at a Solution

a. I've done this part; it's a simple multivariate hypergeometric distribution.
b. This is what confuses me. When you're not including all variables, wouldn't this be called a MARGINAL distribution--so what is the joint distribution? If it said "marginal distribution" I could do that by summing over the possibilities for red.
And isn't this be the same as (a), since when you've found the number of black and white balls, the number of red balls is fixed by $n$...
I'm very confused, thanks for your help!

 

[Ray Vickson]

Homework Statement

An urn contains $p$ black balls, $q$ white balls, and $r$ red balls; and $n$ balls are chosen without replacement.
a. Find the joint distribution of the numbers of black, red, and white balls in the sample.
b. Find the joint distribution of the numbers of black and white balls in the sample.

Homework Equations

The Attempt at a Solution

a. I've done this part; it's a simple multivariate hypergeometric distribution.
b. This is what confuses me. When you're not including all variables, wouldn't this be called a MARGINAL distribution--so what is the joint distribution? If it said "marginal distribution" I could do that by summing over the possibilities for red.
And isn't this be the same as (a), since when you've found the number of black and white balls, the number of red balls is fixed by $n$...
I'm very confused, thanks for your help!

You have it exactly right: the answers to a) and b) are the same. That is true because there are only three colours; if there were 4 or more colours it would not be true; can you see why?