Elevator stop, other approach There is a $10$-floor building, $10$ people get in the elevator in the ground floor and each gets off at one of the floors randomly and independently. What is the probability that the elevator stops at floor $5$?
$$1 - (9/10)^{10}$$
I understand the approach of using the probability that no one gets out on the $3$rd floor, which is $(9/10)^{10}$.
But, my question is how do I approach this problem using the probability that people gets out on the $3$rd floor?
I thought if one person chooses the $5$th floor the probability is $1/10$, and if $2$ people choose the $5$th floor then the probability is $1/9$, and so on. 
Thank you.
 A: Yes, there is an alternative way to solve this, although it's more difficult than the method you mentioned.
Start by considering the case where there are 2 floors and 2 people.  The probability of at least one person exiting on our chosen floor, say floor 2, is
$$P(\geq\text{1 person on floor; 2 people and 2 floors}) = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2}\right) = \frac{3}{4}$$
To see where this came from, we first consider where person 1 gets off.  There is a $1/2$ chance the person will exit on our chosen floor, so we add that to our probability.  In the event that she doesn't get off on floor 2 (an event with probability $1/2$), we consider the other person.  Person 2 has a $1/2$ probability of exiting on floor 2, so we multiply that by the probability that the second person even matters (ie. the probability of the first person not getting off on floor 2).
For 3 people and 3 floors, we'd have
$$P(\geq\text{1 person on floor; 3 people and 3 floors}) = \frac{1}{3} + \frac{2}{3}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{3}\right)\right) = \frac{19}{27}$$
since there's a $1/3$ chance of the first person getting off on our chosen floor, and a $2/3$ probability that they don't and that we have to look at the second person.  The same logic is applied to the second person and so on. 
For 4 people and 4 floors, we'd have
\begin{align}P(\geq\text{1 person on floor; 4 people and 4 floors}) &= \frac{1}{4} + \frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}\right)\right)\right) \\
 &= \frac{175}{256}\end{align}
For $N$ people and $N$ floors, we'd have
\begin{align}&P(\geq\text{1 person on floor; N people and N floors}) \\ =& \frac{1}{N} + \frac{N-1}{N}\left(\frac{1}{N}+\frac{N-1}{N}\left(\;\;\;\;\left.\ldots+\frac{N-1}{N}\left(\frac{1}{N}\right)\right)\right)\ldots\right)
\end{align}
where the pattern continues until there are $N$ of the fractions $1/N$ present in the expression.
By multiplying this out, we have
\begin{align} P =& \frac{1}{N}+\frac{\left(N-1\right)}{N^2}+\frac{\left(N-1\right)^2}{N^3}+\ldots+\frac{\left(N-1\right)^{N-1}}{N^N} \\
=& \sum_{k=0}^{N-1} \frac{1}{N} \left(\frac{N-1}{N}\right)^k
\end{align}
and using the formula for the sum of a finite geometric series, ie
$$\sum_{k=0}^{n-1} a r^k = \frac{a\left(1-r^n\right)}{1-r}$$
we get, 
\begin{align}P =& \frac{\left(\frac{1}{N}\right) \left(1-\left(\frac{N-1}{N}\right)^N\right)}{1-\left(\frac{N-1}{N}\right)} \\
=& 1 - \left(\frac{N-1}{N}\right)^N
\end{align}
which is the same expression as the one given by the more simple method.
