Conditional Expectation in Poisson Distribution So I am currently in the process of learning basic distributions of random variables and I have been trying to understand the following scenario (distilled for sake of brevity):

The number of people who enter an elevator on the ground floor is a Poisson random variable with mean $10$. If there are $N$ floors above the ground floor and if each person is equally likely to get off at any one of these $N$ floors, independently of where the others get off, compute the expected number of stops that the elevator will make before discharging all of its passengers.

So, my thought process is as follows.
Let $X=$ number of people who enter the elevator. $X \sim Poisson(10)$.
Let $Y=$ the number of stops it takes in total.
I define the indicator variable $I_n$ such that it equals $1$ if the elevator stops at a given floor and $0$ otherwise.
Then, $Y = I_1 + ... + I_N$. 
From what I understand, the goal is to find $E[Y]$ since this essentially gives the expected number of times the elevator will stop.
Now, $E[Y] = \sum_{n=1}^N E[I_n] = NE[I_n]$.
Thus, I need to find $I_n$ but this is where I am stuck. 
I have attempted to condition on $X$ (the number of people who entered the elevator) since naturally, the number of stops it takes depends on the number of people that entered to begin with, but then I am not sure how to go after defining $I_n$ in terms of the conditional expectation: $E[I_n] = E[E[I_n | X = m]]$. I know that I need to find $E[I_n | X = m]$ somehow, but I am not sure how to do so.
This is not a homework question and I am really trying to understand the properties of expectation and how to work with distributions on a deeper level rather than just memorizing a strategy for solving these problems and I would greatly appreciate any help.
 A: The number of people getting off on a given floor is a Poisson variable with mean $\frac{10}N$. Thus the probability of not stopping at a given floor is $\mathrm e^{-10/N}$, and the expected number of stops is $N\left(1-\mathrm e^{-10/N}\right)=10-\frac{50}N+O\left(N^{-2}\right)$.
A: Here are some details for @joriki's answer. Let $X_i$ be the number of people getting off at floor $i$, and let $X$ be the total number of people. Then one has to argue that $X_i$ is a Poisson random variable (you can search for "thinning a Poisson process"). We have $X = \sum_{i=1}^N  X_i$. Hence, 
$$ 10 = E[X] = \sum_{i=1}^N E[X_i].$$
By symmetry, all $E[X_i]$ are equal, so $10 = 
N E[X_j]$ that $E[X_j] = 10/N =: \lambda$ for all $j$.
The total number of stops is $Y := \sum_{i=1}^N 1\{X_i \neq 0\}$ (these are the indicators $I_i$ in your notation), hence
$$
E[Y] = \sum_{i=1}^N P(X_i \neq 0) = N P(X_1 \neq 0) = N (1-e^{-\lambda})
$$
using linearity of the expectation and the fact that the expectation of an indicator is the probability of it being 1.
A: This problem is in A First Course in Probability by Sheldon Ross. Given the section in which it occurs, the problem seems designed to have the student work with computing expectations via conditioning. Here is a solution that I believe to be in line with what the author intended, and which is doable by a student who has reached that point in the text.
Using the notation in your question, if $A_n$ is the event that the elevator stops on the $n$-th floor, then
$$
E[I_n \mid X = m] = P(A_n \mid X = m)
  = 1 - P(A_n^c \mid X = m).
$$
For $P(A_n^c \mid X = m)$, we need all $m$ people to not get off on the $n$-th floor, so we have
$$
P(A_n^c \mid X = m) = \left(\frac{N-1}{N}\right)^m.
$$
Putting it together,
$$
E[Y \mid X = m] = N\left(1 - \left(\frac{N-1}{N}\right)^m\right).
$$
Hence,
\begin{align}
E[Y] &= \sum_{m=0}^\infty P(X = m)E[Y \mid X = m]\\
&= \sum_{m=0}^\infty e^{-10}\frac{10^m}{m!} N\left(1 - \left(\frac{N-1}{N}\right)^m\right)\\
&= Ne^{-10}\sum_{m=0}^\infty\frac{10^m}{m!}
  - Ne^{-10}\sum_{m=0}^\infty\frac1{m!}\left(\frac{10(N-1)}N\right)^m\\
&= N - N\exp\left(-10 + 10\left(1 - \frac1N\right)\right)\\
&= N(1 - e^{-10/N}).
\end{align}
After students are comfortable with this kind of reasoning, I usually encourage them to become comfortable with conditioning on the random variable $X$ rather than the event $\{X = m\}$, and to use the so-called "tower property" of conditional expectation. In this problem, it would look like $E[I_n\mid X]=1-P(A_n^c\mid X)$, followed by
$$
P(A_n^c \mid X) = \left(\frac{N-1}{N}\right)^X,
$$
and
$$
E[Y \mid X] = N\left(1 - \left(\frac{N-1}{N}\right)^X\right),
$$
and then finally compute $E[Y]=E[E[Y\mid X]]$. The final computations are the same, it's just a different way of conceptualizing the problem. In other problems, this alternative way of thinking about it can make short work of an otherwise tedious problem.
