Let $X_n$ be a random variable series starting from $n=1$ to $\infty$
We know that $X_n$~$Bin(n, \frac 7 n)$ for all $n>7$.
What is the limit of $\lim _{n\to \infty }\mathbb{P}\left[X_n\le 1\right]=?$
The answer is $8\cdot e^{(-7)}$.
I know how to find the solution non formal, but I want a formalic way, for me.
I tried using Central Limit Theorem but without success, I received $\frac 1 2$ and its not good.
The basic solution is to define $X_n=0$ and $X_n=1$, then sum them, use the limit, Euler limit and we will receive the correct answer.
But this solution is no good for me as I want a formal way with CLT or LLN since the question came when we learned this.
You can use the Law or Rare Events/Poisson Limit Theorem; generally, if $X_n\sim Bin(n,p)$ where $p$ is fixed, then CLT applies. However, since $p=p(n)$ depends on $n$ and $p(n)\to 0$, CLT will not apply, but since $np(n)\to\lambda$ with $0<\lambda<\infty$, we can using another limit theorem as mentioned.
You can image that if $p$ is fixed, then $X_n$ will be approximately normally distributed, but since $p=p(n)\to 0$, this does not hold, and instead $X_n$ is approximately Poisson distributed!
More specifically, the law of rare events says that $X_n$ converges towards a $Pois(\lambda)$-distribution as $n\to\infty$, where $\lambda=\lim_{n\to\infty}np(n)=7$. So $$ \lim_{n\to\infty}\mathbb P(X_n\leq 1)=\lim_{n\to\infty}(\mathbb P(X_n=0)+\mathbb P(X_n=1))=e^{-7}+7e^{-7}=8e^{-7}. $$