An urn contains $100$ blue balls, $100$ yellow balls, $100$ orange balls, $100$ green balls and $100$ red balls. $60$ balls are picked (with replacement) from the urn at random. Let $$ be the number of times a red ball was picked, and let $$ be the number of times a green ball was picked.
The question is split into two parts.
a. What is the joint density of (, )?
b. What is Cov($X,Y$)?
I thought I'd do something like this.
For part (a) $$ \left(x,y\right)\:=\:\frac{60!}{x!y!\left(60-x-y\right)!}\left(\frac{1}{5}\right)^{x+y}\left(\frac{3}{5}\right)^{60-x-y}$$
And for part (b)
$$ cov\left(X,Y\right)\:=\:cov\left(X,60-X\right)=cov\left(X,60\right)+cov\left(X,-X\right)$$ $$=0\:-cov\left(X,X\right)\:=\:-var\left(X\right)=0.96 $$ using this: $$ cov\left(X,60\right)\:=E\left[60X\right]-\:E\left[60\right]E\left[X\right]=60E\left[X\right]-60E\left[X\right]=0 $$ and the fact that X+Y should be equal to 60.
But the answer is: , ~(60, 1/5) and +~(60, 2/5) $$ cov\left(X,Y\right)=\:\frac{var\left(X+Y\right)-var\left(X\right)-var\left(Y\right)}{2}=\frac{60\cdot \frac{1}{5}\cdot \frac{4}{5}-2\cdot 60\cdot \frac{1}{5}\cdot \frac{4}{5}}{2}=-2.4 $$
Can someone help me understand why my method doesn't work?
Thanks!
Community wiki answer so the question can be marked as answered:
As discussed in the comments, $\operatorname{cov}(X,Y)=\operatorname{cov}(X-60-X)$ would hold only if only red and green balls could be picked, no orange or yellow balls.