double posted already but no one would answer so

This constraints represent a sphere of radius 1 inside a cube of side length 2.

Volume of cube  $Vc=2^3=8$

Volume of sphere $Vs=\frac{4}{3}\pi 1^3=\frac{4}{3}\pi$

Probability $p=\frac{Vs}{Vc}$

I'll let you finish!

Try and imagine it.

   $-1\le x\le1\\ -1\le y\le1\\ -1\le z\le1\\$

this encloses a cube.  The centre is (0,0,0)  what is the side length,  What is the area?

Now

$x^2+y^2+z^2\le 1$

is the formula for a sphere with centre (0,0,0) and radius  ?    What is the area?

The sphere is inside the cube.  What fraction of the cube space does the sphere take up?  

That will be your answer.

Just as Alan has already said.