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1989 Paper 2 Q15
D: 1600.0 B: 1484.0
geometric probability expected value semicircle integration triangle area conditional expectation probability density function random points

Two points are chosen independently at random on the perimeter (including the diameter) of a semicircle of unit radius. What is the probability that exactly one of them lies on the diameter? Let the area of the triangle formed by the two points and the midpoint of the diameter be denoted by the random variable \(A\).

  1. Given that exactly one point lies on the diameter, show that the expected value of \(A\) is \(\left(2\pi\right)^{-1}\).
  2. Given that neither point lies on the diameter, show that the expected value of \(A\) is \(\pi^{-1}\). [You may assume that if two points are chosen at random on a line of length \(\pi\) units, the probability density function for the distance \(X\) between the two points is \(2\left(\pi-x\right)/\pi^{2}\) for \(0\leqslant x\leqslant\pi.\)]
Using these results, or otherwise, show that the expected value of \(A\) is \(\left(2+\pi\right)^{-1}\).

Solution:

  1. TikZ diagram
    \begin{align*} \mathbb{E}(A \mid \text{exactly one point on diameter}) &= \int_{-1}^1\int_0^\pi \frac12 (x-0)\cdot 1 \cdot \sin(\pi - \theta) \frac{1}{\pi} \d \theta \frac{1}{2} \d x \\ &= \int_{-1}^1\frac1{2\pi} x \d x \cdot \left [ -\cos \theta \right]_0^\pi \\ &= \frac{1}{2\pi} \end{align*}
  2. TikZ diagram
    \begin{align*} \mathbb{E}(A \mid \text{no point on diameter}) &= \int_0^{\pi} \frac12 \cdot 1 \cdot 1 \cdot \sin x \cdot 2(\pi - x)/\pi^2 \d x \\ &= \frac1{\pi^2} \int_0^\pi \sin x (\pi - x) \d x \\ &= \frac1{\pi^2} \int_0^\pi x\sin x \d x \\ &= \frac1{\pi^2} \left [ \sin x - x \cos x \right]_0^{\pi} \\ &= \frac{1}{\pi} \end{align*}
If both points lie on the diameter the area of the triangle is \(0\). Therefore: \begin{align*} \mathbb{E}(A) &= \frac{1}{2\pi} \mathbb{P}(\text{exactly one point on diameter}) + \frac{1}{\pi}\mathbb{P}(\text{no points on diameter}) \\ &= \frac1{2\pi} \cdot \left (2 \cdot \frac{2}{2+\pi} \cdot \frac{\pi}{2+\pi} \right) + \frac{1}{\pi} \cdot \left ( \frac{\pi}{2+\pi} \cdot \frac{\pi}{2+\pi}\right) \\ &= \frac{1}{\pi} \frac{2\pi + \pi^2}{(2+\pi)^2} \\ &= \frac{1}{2+\pi} \end{align*}

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