How can I calculate the probability that one random variable is bigger than a second one? I have five random variables which are independent and each one of them has a continuous uniform distribution on the interval $ [0,2]:$
$$X_i = \operatorname{Uniform}[0,2].$$
I want to calculate the probability $$\Pr(\min(X_1, X_2, X_3)\gt \max(X_4, X_5)).$$
I'm aware there is combinatorial solution, but I'm trying to solve this problem using coordinates with $X$ as the minimum and $Y$ as the maximum, but I don't know how to sketch the function and calculate the function space in order to know the probability.

For all input values $0 \leq t \leq 2$ we have,
\begin{align*}
\mathbb P (\min(X_1,X_2,X_3) > t ) &= \mathbb P(X_1>t)^3 \\[12pt]
&= \Big (1 - \frac{t}{2} \Big )^3 \\[12pt]
\mathbb P (\max(X_4,X_5) \leq t ) &= \mathbb P(X_4\leq t)^2 \\[12pt]
&= \frac{t^2}{4}, \\[6pt]
\end{align*}
which also gives the corresponding density,
\begin{align*}
\quad f_{\max(X_4,X_5)}(t) &= \frac{t}{2}. \\[6pt]
\end{align*}
Thus, using the substitution $r = t/2$, we have,
\begin{align*}
\mathbb P (\min(X_1,X_2,X_3) > \max(X_4,X_5 )) 
&= \int \limits_0^2 \mathbb P (\min(X_1,X_2,X_3) > t ) \frac{t}{2} \ dt \\[6pt]
&=\int \limits_0^2 \Big (1 - \frac{t}{2} \Big )^3 \frac{t}{2} \ dt \\[6pt]
&= 2 \int \limits_0^1 (1-r)^3 r \ dr \\[6pt]
&= 2 \int \limits_0^1 (r - 3r^2 + 3r^3 - r^4) \ dr \\[6pt]
&= 2 \Bigg[ \frac{r^2}{2} - r^3 + \frac{3r^4}{4} - \frac{r^5}{5} \Bigg ]_{r=0}^{r=1} \\[12pt]
&= 2 \Bigg[ \frac{1}{2} - 1 + \frac{3}{4} - \frac{1}{5} \Bigg ] \\[12pt]
&= 0.1 \\[6pt]
\end{align*}
