The urn problem but with a modification (How does that change the solution?)

Question

These problem is solved in one hand, but then one can ask the following:

Question

Imagine that you have $N$ urns and $b$ balls, you throw the balls and each of them gets into a urn with a probability $p$ or bounce and gets outside the urn with probability $1-p$ so then if we define $$X=\textit{number of filled urns}\;$$ which will be now $E[X]$?

I suspect, by analogy that

$$E[X]=N(1-e^{bp/N})$$

The thing is that I don't know how if my guess is right and how to figure out it.

Attempts:

1) I was trying the following, one defines the random variable,

$$X_{(n)}=\text{number of occupied urns when we throw $n$ balls}$$

then we get that

$$X_{(1)}=1$$

so $X_{(2)}$ could be either $1$ or $2$, so in general one gets that

$$X_{(n)}= X_{(n-1)}+1 \; \text{with probability } \; \frac{X_{(n-1)}}{N}$$

or

$$X_{(n)}= X_{(n-1)} \; \text{with probability } \; \frac{N-X_{(n-1)}}{N}$$

Then we get the recurrence formula:

$$E[X_{(n)}]= (X_{(n-1)}+1)(\frac{X_{(n-1)}}{N})+ (X_{(n-1)})(\frac{N-X_{(n-1)}}{N})$$

But I don't know how to proceed from this.

2) Another thing that can be done is using the above idea but with the conditional expectation, the thing is that I don't know if is a good way of doing it, since we don't know anything about the distribution functions of the random variables $X_{(n)}$.

3) Doing an analogy to what was post in my previous question one only has to modify the probability of being in any urn to the probability of being thrown in any urn, this is

$$(\frac{N-1}{N})^{b} \to (\frac{N-1}{N})^{bp}$$

Am I right?.

Thanks a lot in advance for your help.

Answer 1

Hint: The easiest way is to compute the probability that one specific urn is occupied. The chance that one ball goes in is ???. The chance that one ball doesn't go in is ???. The chance that none of the balls goes in is ???. The chance the urn is occupied is ???. This is the expectation of the variable that is $1$ when the urn is occupied and $0$ otherwise. Now by the linearity of expectation, the expected number of occupied urns is ??? The fact that some balls miss just reduces the effective number of balls to $pb$