Why was poisson distribution introduced? I am studying probabilites and the notion of poisson random variable was introduced in the class. But it seems to me that the introduction of poisson random variable is to provide a easy approximation of the binomial random variable conditioned that n is large and p is small. Besides, the preconditions in the poisson distribution that events are independent of each other seem to come from the fact that binomial random variable is composed by many independent bernouli variables.
So I wonder if originally, poisson distribution was invented to model binomial distribution or was it invented to solve a particular kind of problem
 A: The history of Poisson distribution https://en.wikipedia.org/wiki/Poisson_distribution#History shows that Siméon Denis Poisson introduced the distribution when discussing wrongful convictions of prisoners in a given country by focusing on certain random variables that count the number of discrete occurrences of that take place during a time interval of a given length.
It has since been used for reliability engineering.
To your point, it seems it was generally introduced as a way to model a specific phenomenon but was then applied to solve applications years later.
A: The Poisson is derived from the Binomial distribution, when one takes the limit of n, i.e. the number of experiments, to infinity, while demanding that the expected value stays the same (essentially reducing the probability to 1/n).
Intuitively, it means that there is no confined number of dichotomic (1 or 0) events, but possibly infinite amount of events.
This page explains it nicely (cached version):
Denote E(x) of the binomial distribution, np, by λ.  Note: $p=\frac{\lambda}{n}$
If you develop the Binomial distribution formula, you arrive finally to this:
$P(x) = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left( 1 - \frac{\lambda}{n} \right)^n \left( 1 - \frac{\lambda}{n} \right)^{-x}$
If you take the limit of this expression when n goes to $\infty$, all the first x expressions reduce to 1, and the last expression as well; the second to last becomes $e^{-\lambda}$, and you are left with the Poisson distribution:
$\lim_{n \rightarrow \infty} P(x) = \frac{e^{-\lambda} \lambda^x}{x!}$
Edit: I'm not sure I understand the comments made to this answer. I try to write it more "rigorous-ly":
Suppose I have a series of R.V., s.t. $X_i\sim Bin(n, \frac{\lambda}{n})$. We see that under this setup, the expected value = $n \cdot \frac{\lambda}{n} = \lambda$ stays the same. If I then look at the PMF of this R.V. in the limit, I get that:
$$ \lim P_{X_n}(k) = \lim {n \choose k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k}=\frac{\lambda^k}{k!} \lim \frac{n!}{(n-k)!n^k} (1-\frac{\lambda}{n})^n (1-\frac{\lambda}{n})^{-k}
$$
The 1st and 3rd terms go to 1. The 2nd term goes to $e^{-\lambda}$.  And we get exactly the Poisson PMF.
Hope this clears it up for you guys.
