$0<x<y<z<1, P(z-x<1/3)=$? 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 $0&lt; X &lt; Y &lt; Z &lt; 1$. 
What is the probability of $(Z - X &lt; 1/3)$?

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 $0 &lt; X , Y, Z &lt; 1$.

And two conditions are given

$$0 &lt; X &lt; Y &lt; Z &lt; 1,$$

$$Z-X &lt;1/3.$$

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 $P(z-x&gt;1/3)$.
Then $1/3 &lt; z &lt; 1$, $0 &lt; x &lt; z-1/3$, and $x&lt;y&lt;z$.
I think it is easier to find $P(z-x&gt;1/3)$.
Then $1/3 &lt; z &lt; 1$, $0 &lt; x &lt; z-1/3$, and $x&lt;y&lt;z$.