I'm trying to solve a question about 'order statistics'. $X, Y, Z$ follow uniform distributions between 0 to 1. $(X , Y, Z \sim U(0,1))$
An unequality is given that $0< X < Y < Z < 1$. What is the probability of $(Z - X < 1/3)$?
I can easily calculate the density function of $X, Y, Z$ that $f(x, y, z) = 6.$
I think I should solve a triple integral problem. A cube is given that $0 < X , Y, Z < 1$.
And two conditions are given
$$0 < X < Y < Z < 1,$$
$$Z-X <1/3.$$
Is it a right way to solve this question? I'm a bit of a dummy so a thorough explanation would be much appreciated. Thanks in advance!
I think it is easier to find $P(z-x>1/3)$.
Then $1/3 < z < 1$, $0 < x < z-1/3$, and $x<y<z$.