From [1;2] continuous interval we choose 3 numbers randomly. Let $X$ be the minimum between those numbers.
Find PDF and Expected value.
I fail to understand the problem, since I believe that probability of a precise number to be chosen out of continuous interval is 0. I saw somewhere that it could be solved by CDF $= (2-x)^3$, but I don't know how we get it.
I'd be glad, if you could provide some explanation for a beginner.
probability of a precise number to be chosen out of continuous interval is 0
You're right, but what you says is probability mass function (PMF), and what the question says is probability density function (PDF).
For example, $\Pr(X = 1.5) = 0$ but $\Pr(1.5 < X \leq 1.6) > 0$ and $\Pr(1.5 < X \leq 1.5 + \Delta x) > 0$. Moreover, $\Pr(1.5 < X \leq 1.5 + \Delta x)$ is roughly proportional to $\Delta x$ and $\lim_{\Delta x \to 0} \Pr(1.5 < X \leq 1.5 + \Delta x) / \Delta x$ (probability mass / range) is called probability density.
We'll derive by CDF, because definition of CDF contains a “$\leq$”, which is suitable for describing the minimum.
CDF of each number is
$$ F(x) := \Pr(X \leq x) = \begin{cases} 0 & x < 1. \\ x-1 & 1 \leq x < 2. \\ 1 & 2 \leq x. \end{cases} $$
Now can you follow Henry's comment?
(It's not necessary to fully understand PDF for calculating CDF)
