Why probability for an atom is fraction for population?

In the canonical ensemble the probability of finding an atom (or molecule) in its $i$-th energy state is given by the Boltzmann factor.


This probability is interpreted as the number of microstates that we will find the atom in its $i$-th energy state (neglecting degeneracy) divided by the total number of microstates.


Why we can convert this number to the fraction of atoms (or molecules) that are in this energy state [Wikipedia](https://en.wikipedia.org/wiki/Boltzmann_distribution)?



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> The Boltzmann distribution is often used to describe the distribution of particles, such as atoms or molecules, over bound states accessible to them. If we have a system consisting of many particles, the probability of a particle being in state i is practically the probability that, if we pick a random particle from that system and check what state it is in, we will find it is in state i. This probability is equal to the number of particles in state i divided by the total number of particles in the system, that is the fraction of particles that occupy state i.
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**Is this fraction constant at every time or is the expectation value of the fraction that equals $p\_i$?**


Intuitively we could say, assuming particles behave independently, that each behaves like a toss coin. If we think of each microstate as flipping all the atoms at the same time, then a microstate is like flipping a coin $N$ times, and because $N$ is so large, then the fraction equals $p\_i$.


Also one can view this flipping from a different perspective. Each microstate, is like flipping a number of $N$ coins at the same time, but each flip of the $N$ coins is a **single random experiment**. In this case how can we equate the probability with the fraction of population?


Is there any formal derivation which shows why $p\_i$ equals:
$$ \left\langle \frac{n\_i}{N} \right\rangle $$
?


