Three points on the circumference of a circle are chosen at random. What is the probability the points are all on the same semicircle?
Four points on the surface of a sphere are chosen at random. What is the probability the points are all on the same hemisphere?
Let A and B be random points on the circle with Aa and Bb being diameters. The minor arc ab is common to the semicircular arcs Aba and Bab therefore the probability that a further random point, C, lies on the minor arc ab is
1/2 x 1/2 = 1/4 .
For C to lie on the same semicircle as A and B, it must lie on the major arc ab, so the probability of that happening is 1 - 1/4 = 3/4.
Let A, B and C be random points on the sphere, with Aa, Bb and Cc being diameters. The spherical (minor) triangle abc is common to the hemispheres abc, bca and cab (where the notation abc represents the hemisphere cut off by the great circle through a and b and containing the point c, etc), therefore the probability that a further random point, D, lies on this triangle is
1/2 x 1/2 x 1/2 = 1/8.
For D to lie on the same hemisphere as A, B and C, it must lie outside the triangle abc, so the probability of that happening is 1 - 1/8 = 7/8.
Edited on November 3, 2009, 3:37 pm
Posted by Harry
on 2009-11-03 15:26:54