Solution
The Midpoint Chord
Show the problem again
A point M is chosen uniformly at random inside a disk of radius 1, and the chord of the disk whose midpoint is M is drawn. (If M is the centre, which happens with probability 0, any diameter will do.) What is the expected length of the chord? Give a reduced fraction.
Worked solution
The answer is 4/3. The chord with midpoint M is perpendicular to the radius through M, so if M is at distance r from the centre its length is 2√(1 − r²). For a uniform point in the unit disk, r has density 2r on [0, 1]. Hence E[length] = ∫₀¹ 2√(1 − r²) · 2r dr = 2∫₀¹ √u du = 4/3, using u = 1 − r². This is the "random midpoint" sampling scheme from Bertrand's paradox; it favours short chords because most of the disk's area is far from the centre.
Source: Classic geometric-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.