Complicated probability question with urns and balls

[zeion]

Homework Statement

Suppose that Joe draws k balls from and urn containing n red balls and n green balls, without replacing the balls after they are drawn. Similarly, Mary draws k balls from an urn containing m red balls and m green balls, without replacing the balls after they are drawn. We want to computer the probability that Joe and Mary will draw the same number of red balls.

Homework Equations

The Attempt at a Solution

Let E be the event that J and M draw the same number of red balls.
So P(E) = P(J draws i red balls and M draws i red balls)
= P(J draws i red balls) P(M draws i red balls)

I don't know how to write P(J draws i red balls)

 

[Ray Vickson]

Homework Statement

Suppose that Joe draws k balls from and urn containing n red balls and n green balls, without replacing the balls after they are drawn. Similarly, Mary draws k balls from an urn containing m red balls and m green balls, without replacing the balls after they are drawn. We want to computer the probability that Joe and Mary will draw the same number of red balls.

Homework Equations

The Attempt at a Solution

Let E be the event that J and M draw the same number of red balls.
So P(E) = P(J draws i red balls and M draws i red balls)
= P(J draws i red balls) P(M draws i red balls)

I don't know how to write P(J draws i red balls)

Look up the hypergeometric distribution. See, eg.,
http://en.wikipedia.org/wiki/Hypergeometric_distribution or
http://stattrek.com/lesson2/hypergeometric.aspx .

RGV

 

[zeion]

the chance that J will draw i red ball is

(n choose i) * (n choose k - i) / (2n choose k)

is that right

 

[Ray Vickson]

the chance that J will draw i red ball is

(n choose i) * (n choose k - i) / (2n choose k)

is that right

Yes.

RGV

 

[zeion]

so J and M both draw i balls is

[(n choose i) * (n choose k - i) / (2n choose k)] * [(m choose i) * (m choose k - i) / (2m choose k)]

right