Take the unit square and inscribe a circle $Q$. Then take one of the corners (say the upper-left one) and inscribe a circle $Q'$ in that space (so the circle touch the square twice and $Q$ once). Let $(X_n,Y_n)$ be uniformly distributed on $[0,1]^2$. What is $\mathbb P((X_1,Y_1)\in Q')$?
My motivation for this question is of course the geometrical one, what is the area of said circle. This is the method I came up with, but I am not sure how to proceed.
Adding to Matti's comment. Starting out from the same symmetric top quarter, you know that we have a right triangle, with the corner angles being at $45^\circ$. Now go to the center of the smaller inscribed half-circle and draw a line from the center to the intersection with the square. You can now state an equation about $\sin 45^\circ$ in terms of the small radius $a$.
The equation should look something like this. (For simplicity, I have assumed that the radius of the large circle is 1)
\begin{eqnarray} \frac{1}{\sqrt{2}} &=& \frac{a}{\sqrt{2}-(1+a)}\\ a &=& 3 - 2\sqrt{2} \end{eqnarray}
To answer your question, the probability of landing in one small circle on the unit square is \begin{equation} \frac{\pi a^2}{4} = 0.0231 \end{equation}