What is the confidence to be in average distance from center of circle to uniformly-distributed points within it

Question

I saw here that the average distance from the center of the unit circle to a point in that (the points are uniformly distributed) is $2/3$.

I wonder how to calculate the probability that points located at a distance of $2/3 \pm \epsilon$ from the center (depend on $\epsilon$ of course)?

In math term, I want to calculate $P(\Delta(p_0,p)\in[\frac{2}{3}-\epsilon,\frac{2}{3}+\epsilon])$ where $p$ is random point, $p_0$ is the center of circle, $\Delta$ is a distance between the two points.

Thanks!

Answer 1

If the random point is selected according to a uniform distribution, then the probability is just given by the area of the region divided by the total area of the disc, so the result is $$ \frac{\pi(\tfrac{2}{3} + \epsilon)^2 - \pi(\tfrac{2}{3} - \epsilon)^2}{\pi} $$

Answer 2

As this is about number of points, you have to look at the area. There will be more number of points in the outer part of the circle than closer to the center.

Area between $2/3 \pm \epsilon = \pi [(2/3 + \epsilon)^2 - (2/3 + \epsilon)^2] = \frac {8}{3}\pi.\epsilon$

Divide by the area of the full circle which is $\pi$ so you get $\frac {8}{3}\epsilon$ as the probability of a point being between $2/3 \pm \epsilon$ distance from the center.