A point $(x, y)$ inside of a unit circle is picked uniformly at random. Then a radius $r$ is picked at random such that a circle $(x, y, r)$ is inside the unit circle. What is the probability that unit circle's origin is inside of circle $(x, y, r)$?
I'll assume that the unit circle is centered at the point $(0, 0)$.
I tried to fix the $y$ coordinate of the random point to $0$ and then try to think about all possible $x, r$ values. This has lead me to the following inequalities:
$$0 \leq x \leq 1$$
$$0 \leq r \leq 1 – x$$
This corresponds to all radii such that a new circle is inside of unit one. Now, for a new circle to contain $(0, 0)$
$$x \leq r$$
must hold.
This has lead me to calculating the probability of $\frac{1}{2}$ but it seems to be incorrect. Could you help me please?
Best Answer
First we need to find the probability of $x,y$ to be on distance $R$ from the center: CDF is $P(d<R) = R^2$, so PDF is $\omega(R) = 2R$.
Assuming that “radius picked at random” means, uniform distribution from 0 to 1-R, then the probability of $r-R>0$: $$ \int_0^{1/2} \int_R^{1-R}\omega(R)\omega(r)drdR=\int_0^{1/2} \int_R^{1-R}\frac{2R}{1-R}drdR=\int_0^{1/2}\frac{2R(1-2R)}{1-R}dR\\ = \frac32 - \log 4 $$