How do I find a formula for this? Yes this is homework and no I'm not looking for direct answers, just tips or some clairification on how to do this problem since it doesn't make sense to me and my professor didn't explain it very well. It says

Let $U_{(1)},...,U_{(n)}$ be the values of $n$ independent uniform $(0,1)$ random variables arranged in increasing order. Let $0\le x\lt y\le 1$. Find and justify a simple formula for $P(U_{(1)} \gt x$ and $U_{(n)} \lt y)$.

I just really don't know what to do or how to interpret this. Thanks!
 A: Hint: Let our uniformly distributed random variables be $W_1, W_2, \dots, W_n$. One could also call them $U_1, U_2,\dots, U_n$, but that might risk confusion with the $U_{(i)}$.
$U_{(1)}$ is by definition the minimum of the $W_i$, the smallest result we got in our $n$ experiments. And $U_{(n)}$ is the largest result we got in our $n$ experiments. 
The minimum is $\gt x$ and the maximum is $\lt y$ precisely if all the $W_i$ lie in the interval $(x,y)$. What is the probability that $x\lt W_1\lt y$? What is the probability that $x\lt W_i\lt y$ for all $i$?  
A: What you are dealing with is order statistics:
Here we first have $n$ random variables $X_1,X_2,...,X_n\sim \operatorname{iidUnif(0,1)}$, each is uniformly distributed on the interval $(0,1)$. Now these $U_{(1)},U_{(2)},...,U_{(n)}$ are the order statistics of the random variables $X_1,X_2,...,X_n$. That is,  $U_{(1)}$ is the smallest of the $X_i$'s, and $U_{(2)}$ is the second smallest of the $X_i$'s, and so on.
Note that $U_{(1)}$ is the minimum and $U_{(n)}$ is the maximum of all $X_i$'s. Reinterpret the question as a uniform distribution problem in terms of the $X_i$'s. 
This is usually the case for order statistics problems: to reinterpret the problem in terms the original random variables that got ordered.
