Suppose $X_1, X_2, ... X_n$ are independent and uniformly distributed (on $[0,1]$) random variables. Determine $\lim_{n\to\infty} \mathbb{P}(\sum_{i=1}^n X_i \leq \frac{n}{2})$
My thoughts were the following:
I suppose I can say that $$\lim_{n\to\infty} \mathbb{P}\left(\sum_{i=1}^n X_i \leq \frac{n}{2}\right)=\lim_{n\to\infty} \mathbb{P}\left(\frac{1}{n}\sum_{i=1}^n X_i \leq \frac{1}{2}\right)=\lim_{n\to\infty} \mathbb{P}\left(\overline X_n \leq \frac{1}{2}\right)$$ And isn't $\overline X_n$ also uniformly distributed? So the probability equals $\frac{1}{2}$?
No, $\overline{X_n}$ isn't uniformly distributed; but it's distributed symmetrically about $\frac12$, so you can nevertheless conclude that the probability is $\frac12$ even without the limit.