Poisson Distribution? There exists 1000 boxes. These boxes are randomly filled with balls. How many balls are required in order that only 1 in 100 boxes are left empty?
This sounds like a Poisson distribution problem to me unless I am mistaken.
Where the formula for Poisson Distribution is:
$$Pr(x, y) = \frac{\lambda^{x}e^{-\lambda}}{x!}$$
I deduced that $Pr(x, y) = \frac{1}{100}$ since I have a $\frac{1}{100}$ chance of choosing the selection. I also used $\lambda = 1000$ since I have 10 boxes. Where I try to find the value $x$.
However, the answer I got is a decimal below 1. Obviously not right.

My other attempt is to use the Poison Distribution formula for an expected value. Using the same variable predictions as above.
$E|X-\lambda| = 2exp(-\lambda)\frac{\lambda\lfloor\lambda\rfloor+1}{\lfloor\lambda\rfloor!}$
However, x came out to be 1000 this time. Which I doubt as well.

Am I using the right approach?
To be quite honest, it seems to me that the problem is missing some information on how the balls are chosen, but I am reassured otherwise. Help is appreciated.
 A: The question is well formulated, i think. 
It may be formulated more specifically:
We are placing balls randomly into $n=1000$ boxes and continue the procedure of placing balls until $m=10$ boxes remain empty.
Thus the number $r$ of balls is not fixed in advance but is just a random size. We are interested in expected value of $r$ which we denote by $E[r]$
Now, let us remember a classic problem:
$r$ balls is randomly filled in $n$ boxes. Find the probability of finding exactly $m$ boxes empty. 
We need an asymptotic expression of this probability:
$$P(\lambda) = \frac{\lambda^{m}e^{-\lambda}}{m!}$$
where 
$\lambda=ne^{-\frac{r}{n} } ;n\rightarrow \infty $
of which $r(\lambda)=n\ln\frac{n}{\lambda}$
Finally, using the results above we get:
$$E[r]=\int_0^{\infty}r(\lambda)P(\lambda)dx=4556$$
(for $m=10$ and $n=1000)$
Unfortunately, the integral has no closed form solution.
To get the standard deviation of $r$
$$\delta_r =\sqrt {E[r^2]-E^2[r]}$$
we need to compute
$$E[r^2]=\int_0^{\infty}r^2(\lambda)P(\lambda)dx$$
Result: $\delta_r \approx 300$
