Consider an urn containing $c$ distinguishable balls, $\alpha$ of which are red, $\beta$ of which are blue, and $\gamma$ of which are green, and $\alpha+\beta+\gamma=c$. We assume $\alpha,\beta,\gamma>0$. 


We introduce the non-negative, integer-valued random variable $X$, defined as "the number of independent trials (i.e. with replacement) needed to get at least one red ball and at least one blue ball", and the non-negative, integer-valued random variable $Y$, defined as "the number of independent trials needed to get at least one red ball and at least one blue ball and at least one green ball".



> 
> What are the expected values of $X$ and $Y$?
> 
> 
> 


**My attempt:**


The solution for the case of the variable $X$, as suggested [here](https://math.stackexchange.com/a/2770750), is based on the fact that the probability to get at least one red ball and at least one blue ball, in $i$ independent trials, is 


$$
P(L\_{1,1}^{A,B};i)=1-\left(\frac{\alpha+\gamma}{c}\right)^i-\left(\frac{\beta+\gamma}{c}\right)^i+\left(\frac{\gamma}{c}\right)^i.
$$


Since $X$ is a non-negative and integer-valued variable, its expected value $E[X]$can be written as 


$$E[X]=\sum\_{i=0}^{\infty} P(X>i).$$ 


However, $P(X>i)=1-P(L\_{1,1}^{A,B};i)$. Therefore, since we are dealing with the geometric series,
$$
E[X]=\sum\_{i=0}^{\infty}\left[ \left(\frac{\alpha+\gamma}{c}\right)^i+\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\gamma}{c}\right)^i\right]=\frac{c}{\beta}+\frac{c}{\alpha}-\frac{c}{\alpha+\beta}.
$$


My problem here is: Are we correctly considering the fact that the event $L\_{1,1}^{A,B}$ cannot occur at the first trial? 


I would say "yes", and I am generally confident on this result, but I cannot really extend this reasoning to the variable $Y$.


My solution for the variable $Y$, following this suggestion, would be the following: The probability to get at least one red ball and at least one blue ball and at least one green ball, always in $i$ trials, is $$P(L\_{1,1,1}^{A,B,G};i)=1-\left(\frac{\alpha+\gamma}{c}\right)^i-\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\alpha+\beta}{c}\right)^i+\left(\frac{\alpha}{c}\right)^i+\left(\frac{\beta}{c}\right)^i+\left(\frac{\gamma}{c}\right)^i$$.


Since, again, $Y$ is a non-negative, integer-valued random variable, 
$$E[Y]=\sum\_{i=0}^{\infty} P(Y>i).$$ 


And, again, since $P(Y>i)=1-P(L\_{1,1,1}^{A,B,G};i)$, and applying again the property of the sum of the geometric series,
$$
E[Y]=\frac{c}{\beta}+\frac{c}{\alpha}+\frac{c}{\gamma}-\frac{c}{\alpha+\beta}-\frac{c}{\alpha+\gamma}-\frac{c}{\beta+\gamma}.
$$


The problem I mentioned before appears here more evident: what about the case of $i=0,1,2$? Is the index $i$ correctly treated in the expression of $E[Y]=\sum\_{i=0}^{\infty} P(Y>i)$? 


As observed in this other [post](https://math.stackexchange.com/a/2865474), I would say "no". But then I do not understand how to deal also with the index $i$ in the expression $E[X]=\sum\_{i=0}^{\infty} P(X>i)$.


Thanks for your help!


**EDIT:** A further question: 



> 
> Since we can extract at least once a ball of each of the three colors only if we have already extracted at least once two balls of two different colors, is it correct to assess $E[Y]≥E[X]+1$?
> 
> 
> 



Expected values of two non-negative, integer-valued random variables related to an urn problem
For each $j=1,2,3$, let $p\_j$ denote the probability of getting a red, blue, green ball (respectively) in one trial. Let $(p,q,r):=(p\_1,p\_2,p\_3)$, so that $p+q+r=1$. 
For $x=0,1,\dots$ and $j=1,2,3$, let $N\_{x,j}$ denote the number of trials with outcome $j$ among the first $x$ trials. Then for $x=0,1,\dots$
\begin{align}
 P(X>x)&=P(N\_{x,1}=0\text{ or }N\_{x,2}=0) \\ 
 &=P(N\_{x,1}=0)+P(N\_{x,2}=0)-P(N\_{x,1}=0\text{ and }N\_{x,2}=0) \\
 &=(q+r)^x+(p+r)^x-r^x
\end{align}
and similarly
\begin{align}
 P(Y>x)&=P(N\_{x,1}=0\text{ or }N\_{x,2}=0\text{ or }N\_{x,3}=0) \\ 
 &=(q+r)^x+(p+r)^x+(p+q)^x-r^x-q^x-p^x+0^x,
\end{align}
with $0^0:=1$. 


Thus, 
\begin{equation}
 EX=\sum\_{x=0}^\infty P(X>x)=\frac1p+\frac1q-\frac1{p+q}
\end{equation}
and 
\begin{equation}
 EY=\sum\_{x=0}^\infty P(Y>x)=\frac1p+\frac1q+\frac1r-\frac1{q+r}-\frac1{p+r}-\frac1{p+q}+1. 
\end{equation}
It seem that your answer for $EY$ misses the summand $1$.


