Solution
A High Median
Show the problem again
Five numbers are drawn independently and uniformly from [0, 1]. What is the probability that their median exceeds 3/4? Give a reduced fraction.
Worked solution
The answer is 53/512. The median of five numbers exceeds 3/4 exactly when at least three of the five exceed 3/4. Each number independently exceeds 3/4 with probability 1/4, so the count is binomial(5, 1/4): P = C(5,3)(1/4)³(3/4)² + C(5,4)(1/4)⁴(3/4) + (1/4)⁵ = (90 + 15 + 1)/1024 = 106/1024 = 53/512. This translation, from "the k-th order statistic exceeds t" to "at least n − k + 1 samples exceed t", turns most order-statistic tail questions into binomial sums.
Source: Classic order-statistics exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.