I was thinking in the following problem:
Imagine you have $N$ urns and $b$ balls, so you define the random variable $X= \text{number of filled urns}$, then a question arises, How can one know which is the Expected value $E[X]$ of $X$?
I thought this was a new one but someone told me that this was a well known result and in fact that
$$E[X]=N(1-e^{-b/N})$$
So how this could be? I can't figure out why is this true.
Thanks a lot in advance.
Let $X_i$ be the indicator variable for the $i^{th}$ urn. That is, $X_i=1$ if the $i^{th}$ urn is empty, $0$ otherwise.
By Linearity of Expectation we know that the answer $E$ is given by: $$E=E\left[\sum X_i\right]=\sum E[X_i]=Np$$ Where $p$ is the probability that any specified urn is empty. It's easy to compute that $p=\left(\frac {N-1}N\right)^b$ So $$E=N\times \left(\frac {N-1}N\right)^b$$
If $N$ is very large, the standard limit definition of the exponential shows that your formula is a decent approximation.