On a circumference of circle which radius is 1, two points are chosen randomly. What is the probability that the distance between these two points is less then 1?
The solution from my book is $\frac{1}{3}$. I don't have an idea to solve this.
Geometric probability; circle and two points
geometric-probabilityprobability
Related Solutions
The answer is $1-nP = 1-\frac{n}{2^{n-1}}$. You can see this as follows:
Select and fix a direction around the circle (clockwise or counterclockwise). For the $i$-th point $X_i$ selected, what is the probablity that all the other points are not inside the semi-circle started by $X_i$ and in the selected direction? The answer is $P=\frac1{2^{n-1}}$. If that is the case, then the center of the circle does not lie inside the polygon.
Conversely, if the center of the circle does not lie inside the polygon, there must be a pair of 'consecutive' (in the chosen direction around the circle) points who are more than a semi-circle away (in that direction). The first one of the them fills the role of the point $X_i$ above.
To sum up, the center of the circle does not lie inside the polygon iff there exists a point $X_i$ such that all the other points are not inside the semi-circle started by $X_i$ and going in the chosen direction.
We also known the probability for that event, if we chose the index $i$ beforehand: $P=\frac1{2^{n-1}}$
To get the probability that this happens for any $i$, we apply the principle of inclusion and exclusion.
The good thing is the formulas get really easy, because the probability that the event happens for more than one index is zero! That would imply two different non-overlapping (at most touching at the end) segments without any chosen points inside, that are both longer than a semi-circle. That can't happen, of course.
That means the probability that for any index $i$ the point $X_i$ starts an 'empty' semi-circle is just the sum of all the single probabilities, namely $\frac{n}{2^{n-1}}$.
Since you are looking for the opposite event, the probability you seek is $1-\frac{n}{2^{n-1}}$
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 $\frac{2\sin^{-1}d/2}\pi$.
Best Answer
Say the circle is centered at $O$ and fix one point at $P$. Let $A$ and $B$ be the two points on the circle at distance $1$ from $P$. $\angle AOP=60^\circ$ (draw a picture to see the equilateral triangle,) so $\angle AOB=120^\circ$, one-third of the circle.