thanks for the help in advance. I was given an assignment where i have to prove that $f(x,y,z)=24x$ where $0<x<y<z<1$ is a pdf, so I can later calculate $P(X+Y<1)$ and $P(Z-X>\frac 12)$.
I have already proved that this is a joint fdp using $\int_0^1 \int_0^z \int_0^y (24x) dxdydz$ but i can't find how to get $P(X+Y<1)$ or $P(Z-X>\frac 12)$. If there were only 2 variables in my function i'd just have to graph $x+y=1$ to get the limits of integration, but in this case the three variables throw me off. ¿What do i have to do get the limits of integration?
$$\Pr[X+Y < 1] = \int_{z=0}^1 \int_{y=0}^z \int_{x=0}^{\min(y,1-y)} 24x \, dx \, dy \, dz.$$ This is because if $X + Y < 1$, and $0 < X < Y$, this implies $X < 1-Y$ and $X < Y$. So $X < \min(Y, 1-Y)$. When do we take the minimum as $Y$ versus $1-Y$? Obviously, if $Y = 1-Y$, then $Y = 1/2$, so $$\min(Y, 1-Y) = \begin{cases} Y, & Y \le 1/2 \\ 1-Y, & Y > 1/2. \end{cases}$$