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## Homework Statement

Let r,g,b∈ℕ. r red, g green and b black balls are placed in an urn.

The balls are then drawn one at a time with replacement, until a black ball is picked for the first time. (1)

X counts the number of red balls and Y the number of the green ones, until a black one is picked.

Find EX, EY, Var(X), Var(Y) and ρ(X,Y)=cov(X,Y)/σ_Xσ_Y

## Homework Equations

Expectation value: [itex]E(X)=\sum_{i∈I}x_iP(X=x_i)[/itex]

Multinomial distribution for 3 different balls: [itex]P(X=x, Y=y, Z=z)=\frac{n!}{x!y!z!}p_1^xp_2^yp_3^z[/itex], with [itex]n=x+y+z[/itex]

## The Attempt at a Solution

The probabilities for drawing a red ball, [itex]p_1=\frac{r}{r+g+b}[/itex], green [itex]p_2=\frac{g}{r+g+b}[/itex] and black [itex]p_3=\frac{b}{r+g+b}[/itex]

I thought X and Y was i.d.d to the binomial distribution and would therefore have the means [itex]EX=np_1[/itex] and [itex]EY=np_2[/itex] but then the condition (1) isn't used.

I then thought about calculating [itex]E(X)=\sum_{i∈I}x_iP(X=x, Y=y, Z=1)=\sum_{i∈I}x_iP(X=x, Y=y, Z=1)=\frac{(x+y+1)!}{x!y!1!}p_1^xp_2^yp_3[/itex] where I let Z denote the number of black balls drawn.

But I got stuck again, not knowing what variable to sum for, and I think it's wrong to write [itex]E(X)[/itex] because the sum is not only for the random variable X.

Any tips are much appreciated

Alex