Four points are chosen at random on the surface of a sphere. Determine the probability that the center of the sphere lies inside the tetrahedron described by these four points.
I agree with Dustin.
If the points are all in a single hemisphere, then they clearly do not contain the center (except in the zero probability case where three of the points lie on a great circle).
Working in the other direction, it is also pretty clear that if the tetrahedron does not contain the center, then there must be planes which divide the sphere in half and which do not intersect the tetrahedron. Therefore, if the tetrahedron does not contain the sphere's center, then its points all lie in a single hemisphere. Contrapositively, if all of its points are not in a single hemisphere, then the tetrahedron does contain the sphere's center.
Therefore, the center of the sphere lies inside the tetrahedron described by these four points if and only if the four points do not lie in a single hemisphere.