# 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.

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There may be some missing requirement. It is possible, though highly unlikely, that if we assign $100000$ balls at random, they will all end up in the third bin. The problem may ask for the smallest number of balls such that with probability at least 0.95$, at most$1\$ bin will remain empty. – André Nicolas Mar 6 '13 at 5:17
I'm glad that I am not the only one who thinks that this is a poorly designed question. Especially from someone with a lot of reputation. – jakebird451 Mar 6 '13 at 5:43