A-6. Take any four points on a sphere to form a tetrahedron. What is the probability that the center of the sphere is inside the tetrahedron? (It can be assumed that each point is independently chosen and is uniformly distributed on the sphere)
On the sphere, consider three points $P, Q, R$, and we are interested in the position of another point $S$ on the sphere. Point $S$ must be such that the tetrahedron $P Q R S$ contains the center of the sphere $O$. Let $P^{\prime}, Q^{\prime}, R^{\prime}$ be the points symmetric to $P, Q, R$ with respect to the center of the sphere. It is easy to see that point $O$ is inside the tetrahedron $P Q R S$ if and only if $S$ lies within the solid angle formed by the line segments $O P^{\prime}, O Q^{\prime}, O R^{\prime}$. Therefore, if $P, Q, R$ are chosen, the probability that point $S$ satisfies the condition is the area of the spherical triangle $P Q R$ divided by the surface area of the sphere (since the triangles $P Q R$ and $P^{\prime} Q^{\prime} R^{\prime}$ are equivalent).

The probability we need to find is the area of $P Q R$ divided by the surface area of the sphere (where $P, Q, R$ are chosen randomly). We note that the area of $P Q R$ is equal to the area of $P^{\prime} Q R$, and their combined area is the area of the spherical sector with poles at $P, P^{\prime}$ (i.e., the area of the sphere between two great circles—translator's note). This random sector corresponds to a random angle (varying along $P P^{\prime}$) (i.e., the dihedral angle between the planes of the two great circles—translator's note) ranging from 0 to $\pi$. The average area of this random sector is $\frac{1}{2}$ of a hemisphere, so the average area of $P Q R$ is $\frac{1}{8}$ of the sphere's surface. Therefore, the probability that the condition is satisfied is $\frac{1}{8}$.