What is the distribution of $\min(r_1, r_2)$? 
If I have 2 uniform random numbers $r_1$ and $r_2$ in $[0,1)$, what is the distribution of $\min(r_1, r_2)$?

I have a problem I'm trying to get my arms around, and getting some context on this will help.
(BTW, my next step is to look at the minimum of a set of uniform randoms, $r_1$ to $r_n$, but this is my starting point.)
 A: If $r_1, r_2$ $-$ independent random variables with uniform distribution on $[0,1)$, then it is easy to use geometric interpretation of probability.
Universal sample space  $\Omega$ for pair $(r_1,r_2)$ is the square $[0,1) \times [0,1)$.
Cumulative distribution function for random variable $r = \min \{r_1,r_2\}$ is
$F(r_0) = P(r\leqslant r_0) = \displaystyle\frac{{\rm mes}\;A(r_0)}{{\rm mes}\;\Omega}$, 
where $A(r_0)\subset \Omega$ is the subset of $\Omega$, where $\min\{r_1,r_2\}\leqslant r_0$; 
other words, complement of $A(r_0)$ is  $\overline {A(r_0)} = \{(r_1,r_2)\in\Omega: r_1>r_0, r_2>r_0\}$ .
In fact, $\overline {A(r_0)}$ is the square: $(r_0,1)\times (r_0,1)$.
So, ${\rm mes}\:A(r_0) = {\rm mes}\:\Omega - {\rm mes}\:\overline {A(r_0)} = 1 - (1-r_0)^2 = 2r_0 - r_0^2$.
Hence $F(r_0) = \displaystyle\frac{2r_0-r_0^2}{1} = 2r_0 - r_0^2$, if $r_0 \in (0,1)$.
Finally, cumulative distribution function is
$F(r) = \;\;\;\begin{array}{lll} 
0, \;\;{\rm if}\;  r<0; \\ 
2r-r^2, {\rm if}\; r\in [0,1); \\
1, \;\;{\rm if}\; r\geqslant 1. \end{array}$
Probability density function is $f(r) = F'(r)$: 
$f(r)= 2-2r$, if $r\in [0,1)$; and is zero-function outside.

(In the case of $(r_1, \ldots, r_n)$ we'll consider $\Omega$ as $n$-dimensional cube.
${\rm mes} A(r_0) = 1 - (1-r_0)^n$. )
