Random points in a sphere

The volume bounded by $|x|+|y|+|z| < 1$ is an octahedron with vertices at $(±1,0,0)$ , $(0,±1,0)$ , and $(0,0,±1)$ .

So the probability that $|x|+|y|+|z| < 1$ will be given by the volume of the octahedron divided by the volume of the volume of the unit sphere.

$P = \frac{\frac{4}{3}(1^3)}{\frac{4}{3}\pi(1^3)} = \frac{1}{\pi} = \boxed{0.318}$