A circle is randomly chosen in a circle of radius 1 in the sense that a point is randomly chosen for its center, then a radius is chosen at random so that the new circle is contained in the original circle. What is the probability that the new circle contains the center of the original circle?
Solution: If the center of the new circle is more than $1 / 2$ away from the center of the original circle then the new circle cannot possibly contain the center of the original one. Let $x$ be the distance between the centers (by symmetry this is all we need to consider), then for $0 \leq x \leq 1 / 2$ the probability of the new circle containing the center of the original one is $1-\frac{x}{1-x}$. Hence we need to compute $\int_{0}^{1 / 2}\left(1-\frac{x}{1-x}\right) d x=\frac{1}{2}-\int_{0}^{1 / 2} \frac{x}{1-x} d x$. To evaluate the integral, we can integrate by parts to get
$$
-\left.x \ln (1-x)\right|_{0} ^{1 / 2}-\int_{0}^{1 / 2}-\ln (1-x) d x=-\frac{1}{2} \ln \left(\frac{1}{2}\right)-[(1-x) \ln (1-x)-(1-x)]_{0}^{1 / 2}=\ln 2-\frac{1}{2} .
$$

Alternatively, we can use polynomial division to find that $\frac{x}{1-x}=-1+\frac{1}{1-x}$, so $\int_{0}^{1 / 2} \frac{x}{1-x} d x=$ $\int_{0}^{1 / 2}\left(-1+\frac{1}{1-x}\right) d x=\ln 2-\frac{1}{2}$. Therefore the probability is $\frac{1}{2}-\left(\ln 2-\frac{1}{2}\right)=\mathbf{1}-\ln \mathbf{2}$.