Suppose that $Y$ balls are placed randomly in $r$ numerated boxes, where $Y$ is a random variable with Poisson distribution with parameter $\lambda$. Let $X$ be the number of balls in the first box. Determine the distribution of random variable $X$ and calculate $E[X]$.
My attemp:
I need to calculate
\begin{align} P[X=x]&=\sum_{y=0}^{\infty}P[X=x|Y=y]P[Y=y]\\ &=\sum_{y=0}^{\infty} \left( \frac{\binom yx (r-1)^{y-x}}{r^y} \right) \left( \frac{e^{-\lambda} \lambda^y}{y!} \right)\\ &=\sum_{y=0}^{\infty} \left( \frac{(r-1)^{y-x}}{x! (y-x)! r^y} \right) \left( e^{-\lambda} \lambda^y \right)\\ &=\frac{e^{-\lambda}}{x!}\sum_{y=0}^{\infty} \left( \frac{(r-1)^{y-x}}{ (y-x)! r^y} \right) \left( \lambda^y \right)\\ &=\frac{e^{-\lambda}}{x!}\sum_{y=0}^{\infty} \left( \frac{(r-1)^{y-x}r^{-x}}{ (y-x)! r^y r^{-x}} \right) \left( \lambda^y \frac{\lambda^{-x}}{\lambda^{-x}} \right)\\ % &=\frac{e^{-\lambda}r^{-x}}{x!\lambda^{-x}}\sum_{y=0}^{\infty} \left( \frac{(r-1)^{y-x}}{ (y-x)! r^{y-x}} \right) \left( \lambda^{y-x} \right)\\ % &=\frac{e^{-\lambda} \left(\frac{\lambda}{r}\right)^{x} }{x!}\sum_{y=0}^{\infty} \frac{\left(\frac{r-1}{r} \lambda \right)^{y-x}}{ (y-x)!}\\ % &=\frac{e^{-\lambda} \left(\frac{\lambda}{r}\right)^{x} }{x!}\sum_{y=0}^{\infty} \frac{\left(\lambda-\frac{\lambda}{r}\right)^{y-x}}{ (y-x)!}\\ % &=\frac{e^{-\lambda} \left(\frac{\lambda}{r}\right)^{x} }{x!} e^{\lambda-\frac{\lambda}{r}}\\ % &=\frac{\left(\frac{\lambda}{r}\right)^{x} }{x!} e^{-\frac{\lambda}{r}}\\ \end{align}
Hence, notice that $X \sim $ Poisson$(\frac{\lambda}{r})$ and $E[X]=\frac{\lambda}{r}$.
I do not know if this aproach is the best and if $P[X=x|Y=y]$ is correct. Is this correct?