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.