# Multiple Random Variables - find probability given joint pdf

## Homework Statement

Show that the function defined by f(x,y,z,u) = 24*(1+x+y+z+u)^(-5) for x,y,z,u>0 and f=0 elsewhere is a joint density function.

Find P(X>Y>Z>U) and P(X+Y+Z+U>=1).

## Homework Equations

distribution function = quadruple integral from 0 to x (or y or z or u) here of the joint density.
P(X<=x, Y<= y, Z <= z, U <= u) = F(x,y,z,u) = multiple integral (each variable from 0 to infinity) of f(x,y,z,u)

## The Attempt at a Solution

I've checked that f is a joint density function by integrating from 0 to infinity for all variables and getting 1 as an answer.

What I'm unsure of is how to find the probabilities requested. I've also found that the marginal density of X is f_X(x)=(1+x)^(-2) by integrating from 0 to infinity for the other three variables, so F_X(x)=1-1/(x+1).

I think this all has to do with the bounds of the integral of the joint pdf, but I'm having trouble figuring out what they are.
For the first probability, P(X<Y<Z<U), I know from the solutions it's equal to 1/24, and I'm tempted to think that since the marginal densities involve only one variable this can be reduced to essentially a counting problem where there are 4! orders and one that you want so the probability is 1/4! = 1/24. But I'm not sure if this is legitimate or if it's just a coincidence and the actual solution requires integration and such.

For the second probability, P(X+Y+Z+U>=1), I thought it should be the integral of the joint pdf for u,z,y from 0 to 1 (because if they are greater than 1 then P=1 which isn't interesting) and x from 0 to 1-y-z-u but this didn't work out (the solution says the answer should be 15/16).

If anyone could offer some advice I would greatly appreciate it!
Thanks so much!!