Let $X_1,X_2,…$ be a sequence of independent and identically distributed random variables with $P(X_1=1)=\frac{1}{4}$ and $P(X_1=2)=\frac{3}{4}$. If $\bar{X_n}=\frac{1}{n}\sum_{i=1}^{n}X_i$, for $n=1,2,\ldots$ then $\lim_{n\to \infty} P(\bar{X_n}\leq 1.8)$ is ?
My work:
We can write $P(\bar{X_n}\leq1.8)=1-P(\bar{X_n}\geq1.8)$,
so $\lim_{n\to \infty}P(\bar{X_n}\leq1.8)=1-\lim_{n\to \infty}P(\bar{X_n}\geq1.8)=1-P \{\lim_{n\to \infty}\bar{X_n}\geq 1.8\}$.
Now I have a feeling that we can apply Weak law of large numbers, but I don't know the mean. If it were the case then we could conclude that required probability is $1$. So how should I proceed next? Is this the right path? Help please. Thanks.
Since $\overline{X}_n=\frac74$ and $\mathrm{Var}\left(X_n\right)=\frac3{16n}$, we have by Chebyshev's Inequality, $$ P\left(\left|X_n-\tfrac74\right|\ge\lambda\right)\le\frac3{16n\lambda^2} $$ Plugging in $\lambda=\frac1{20}$ yields $$ P\left(X_n\ge1.8\right)\le\frac{75}{n} $$ Therefore, $$ P\left(X_n\le1.8\right)\ge1-\frac{75}{n} $$