We randomly uniformly pick point $x$ on a unit circumference $\{x^2 + y^2 = 1\}$. Then we randomly uniformly pick point $y$ on a unit circle $\{x^2 + y^2 \leq 1\}$. Let $R$ be a rectangle with diagonal $xy$, which sides are parallel to coordinate axes. What is the probability that $R$ lies completely inside the unit circle?
We see, that after $x$ is chosen we have a rectangle $A$, within which our point $y$ should fall to satisfy the condition of the task.
The area of the rectangle $A$ equals $S_A(x) = 4 \sin{(2x)}$ and the probability $P(\{y \in A\}) = \frac{S_A}{S_{circle}} = \frac{S_A}{\pi}.$ How can I find this probability?
The problem for me is that $S_A(x)$ is the transformation of random variable $x \sim unif(0, \frac{\pi}{2})$. For some reason I think that the answer is $\frac{\mathbb{E}[S_A(x)]}{\pi} = \frac{1}{\pi} \cdot \frac{1}{\pi/2 – 0} \cdot \int_0^{\pi/2}S_A(x)dx$, but I do not know why I think so.
Best Answer
For a given position of $X$, say at coordinates $(\cos\theta,\sin\theta)$ in the first quadrant, the area of the rectangle is indeed
$$A=4\sin\theta\cos\theta.$$
As $Y$ is drawn uniformly in the disk, the probability to fall in $A$ is $A/\pi$.
Then as $X$ is drawn uniformly on the circumference and independently of $Y$, you take the expectation on the first quadrant,
$$P=\frac4\pi\frac 2\pi\int_{\theta=0}^{\pi/2}\sin\theta\cos\theta\,d\theta=\frac4{\pi^2}.$$