Understanding the Joint Distribution of Balls in an Urn

  • Thread starter Thread starter michaelxavier
  • Start date Start date
  • Tags Tags
    Multivariate
michaelxavier
Messages
14
Reaction score
0

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!
 
Physics news on Phys.org
michaelxavier said:

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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top