Let $C$ be the unit circle $x^{2}+y^{2}=1$. A point $p$ is chosen randomly on the circumference of $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$-and $y$-axes with diagonal $p q$. What is the probability that no point of $R$ lies outside of $C$?
The pair $(p, q)$ is chosen randomly from the three-dimensional domain $C \times$ int $C$, which has a total volume of $2 \pi^{2}$ (here int $C$ denotes the interior of $C$ ). For a fixed $p$, the locus of points $q$ for which $R$ does not have points outside of $C$ is the rectangle whose diagonal is the diameter through $p$ and whose sides are parallel to the coordinate axes (Figure 112). If the coordinates of $p$ are $(\cos \theta, \sin \theta)$, then the area of the rectangle is $2|\sin 2 \theta|$Figure 112The volume of the favorable region is therefore$$ V=\int_{0}^{2 \pi} 2|\sin 2 \theta| d \theta=4 \int_{0}^{\pi / 2} 2 \sin 2 \theta d \theta=8 . $$Hence the probability is$$ P=\frac{8}{2 \pi^{2}}=\frac{4}{\pi^{2}} \approx 0.405 . $$