Solution
All in a Semicircle
Show the problem again
Three points are chosen independently and uniformly at random on a circle. What is the probability that all three lie within some semicircle? Express as a fraction.
Worked solution
The answer is 3/4. For each point i, let A_i be the event that the semicircle starting at point i and running clockwise contains the other two points. These events are disjoint (at most one point can be the "clockwise-first"), and each occurs with probability (1/2)^2 = 1/4. So the answer is 3 × 1/4 = 3/4. In general, for n points the probability is n/2^(n−1).
Source: Classic geometric-probability exercise found in standard probability texts; no single originator. Statement written for AxiomIQ.