probability distribution , mean and covariance of balls in an urn So I have the following question in "probability":
An urn contains three balls: white, blue and red. At each stage a ball is picked up randomly and, if it is not red, it is returned to the urn. The process is repeated until the first time the red ball is picked up. Then the process ends. Denote by: X the number of times the blue ball was chosen; Y the number of times the white ball was chosen; and L the length of the process. For example, if the blue ball was chosen the first three times, the fourth time the white ball was chosen, and the fifth time the red, then X=3; Y=1; L=5.
I need to fine E(Y) and COV(X,L), but it's not the problem. I'm having troubles in finding the Probability distribution of Y,X and L.
I would be grateful if someone could help me on this, on how to calculate P(Y=k),P(X=k),P(L=k) for each k=1,2,…
thank you very much!
 A: On any pick there is an $1/3$ probability of incrementing $Y$, an $1/3$ probability of incrementing $X$, and an $1/3$ probability of ending the experiment.

$\mathsf P(L=k)$ is the probability that you pick $k-1$ balls before the first red.   This is a geometrical distribution.
$$\begin{align}
\mathsf P(L=k)
 & = \left(1-\frac 1 3\right)^{k-1}\cdot\frac 1 3
\\[1ex] & = \frac{2^{k-1}}{3^k}
\end{align}$$

$\mathsf P(X=k)$ is the probability that you pick $k$ blue balls, and any number of white balls, before the first red.   This is almost a geometrical distribution except for those darn white draws.
But wait.   When counting $X$ we just don't care about turns when a white ball is drawn; we only worry about whether the draw is blue or red, and there is an equal (conditional) probability  that the draw will be either when it is not-white.
That looks like a geometric distribution after all.   What do you think it would be?

$\mathsf P(Y=y)$ is the probability that you pick $k$ white balls, and any number of blue balls, before the first red.
Oh, the symmetry!
