Problem source

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 $A$. 
\begin{questionparts}
\item Given that exactly one point lies on the diameter, show that the expected value of $A$ is $\left(2\pi\right)^{-1}$. 
\item 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 $0\leqslant x\leqslant\pi.$]
\end{questionparts}
Using these results, or otherwise, show that the expected value of $A$ is $\left(2+\pi\right)^{-1}$.

Solution source

\begin{questionparts}
\item 
\begin{center}
    \begin{tikzpicture}
        \draw (-2, 0) -- (2,0);
        \draw (2,0) arc (0:180:2);

        \draw (0,0) -- ({2*cos(30)}, {2*sin(30)}) -- (-1.2, 0);

        \coordinate (A) at (2,0);
        \coordinate (O) at (0,0);
        \coordinate (X) at  ({2*cos(30)}, {2*sin(30)});

        \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = A--O--X};

        \draw[<->] (0,-0.1) -- (-1.2, -0.1);
        \node at (-.6, -0.2) {$x$};
    \end{tikzpicture}
\end{center}
\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*}
\item \begin{center}
    \begin{tikzpicture}
        \draw (-2, 0) -- (2,0);
        \draw (2,0) arc (0:180:2);

        \coordinate (A) at (2,0);
        \coordinate (O) at (0,0);
        \coordinate (X) at  ({2*cos(30)}, {2*sin(30)});
        \coordinate (Y) at  ({2*cos(100)}, {2*sin(100)});

        \draw (O) -- (X) -- (Y) -- cycle;


        \pic [draw, angle radius=0.8cm, "$x$"] {angle = X--O--Y};

    \end{tikzpicture}
\end{center}
 \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*}
\end{questionparts}

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*}