Solution
A Tilted Cap
Show the problem again
A point (x, y, z) is chosen uniformly at random on the surface of the unit sphere x² + y² + z² = 1. What is the probability that x + y + z > 1? Give an exact closed form, or a decimal rounded to 4 places.
Worked solution
The answer is (3 − √3)/6 ≈ 0.2113. Archimedes' hat-box theorem says that the area of a spherical zone between two parallel planes is proportional to the distance between the planes. Equivalently, for a uniform point on the unit sphere, its coordinate along any fixed unit vector is uniform on [−1, 1]. Take the unit vector n = (1, 1, 1)/√3. Then x + y + z = √3 · (point · n), and the condition becomes point · n > 1/√3. Since point · n is uniform on [−1, 1], the probability is (1 − 1/√3)/2 = (3 − √3)/6.
Source: Uses Archimedes' hat-box theorem; problem in wide circulation in this form. Statement written for AxiomIQ.