A point in space $(x,y,z)$ is randomly selected so that $-1\le x \le 1$,$-1\le y \le 1$,$-1\le z \le 1$. What is the probability that $x^2+y^2+z^2\le 1$?
+506 The equation \(x^2+y^2+z^2=1\) represents a sphere with a center at (0, 0, 0) and a radius of 1, and the volume of the space that contains all the possible points of \((x, y, z)\) is a cube with a side length of 2. The volume of the sphere is \(\frac{4}{3}\pi\), and the volume of the cube is \(8\), so the probability that \(x^2+y^2+z^2\leq1\) is \(\frac{\frac{4}{3}\pi}{8}=\boxed{\frac{\pi}{6}}\)
