Imagine we have a unit circle and randomly select two points on the circumference of this circle, say $A$ and$ B$. What is the probability that this distance between the points $A$ and $B$ is less than some fixed value, say $d$, where $d\le2$. Is there maybe some way to set this up using triple integrals? I am honestly not really sure where to start.
Thanks for any input you might have!
Given a unit circle, the distance between two points on it that are an angle of $\theta$ apart can be worked out as $d=2\sin\frac\theta2$, so $\theta=2\sin^{-1}\frac d2$. Arbitrarily fix $A=(1,0)$, then $B$ is less than $d$ away from $A$ iff the absolute angle it makes with the $+x$-axis is less than $2\sin^{-1}\frac d2$, which happens with probability $\frac{2×2\sin^{-1}d/2}{2\pi}$. So the final result is $\frac{2\sin^{-1}d/2}\pi$.
Let's assume that the selection of the two points is independent, and that both probabilities are uniformly distributed on the circle. Then we can fix $A$, say at the bottom, and you only need to deal with $B$. I will do a simple geometrical construction: draw a circle of radius $d$, centered on $A$. Let's call the intersections of the two circles $L$ and $R$ (left and right). Then the probability of getting $B$ at distance smaller than $d$ is the ratio of the bottom arc $LAR$ and the circumference of the original circle
