Solution
Four Points in a Disk
Show the problem again
Four points are chosen independently and uniformly at random inside a disk. What is the probability that one of the four points lies inside the triangle formed by the other three (so that their convex hull is a triangle rather than a quadrilateral)? Give an exact closed form, or a decimal rounded to 4 places.
Worked solution
The answer is 35/(12π²) ≈ 0.2955. Take the disk to have radius 1 and area π. Point D lies inside triangle ABC with probability E[area(ABC)]/π, conditioning on A, B, C. At most one of the four points can lie inside the triangle of the others, so these four events are disjoint and P = 4·E[area of a random triangle in the disk]/π. The expected area of a triangle with three uniform vertices in the unit disk is the classical value 35/(48π), obtainable with Crofton's differential method or the Blaschke-Petkantschin formula. Therefore P = 4·(35/(48π))/π = 35/(12π²), and the four points are in convex position with probability 1 − 35/(12π²) ≈ 0.7045. Among all convex regions, the disk maximises the convex-position probability (Blaschke), while triangles minimise it at 2/3.
Source: Sylvester's four-point problem (J. J. Sylvester, 1864), here for a disk; the value is classical. Statement written for AxiomIQ.