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!
On
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
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 $\frac{2\sin^{-1}d/2}\pi$.