# Probability of a Chord Passing Through Two Concentric Circles

One of my friends gave me a problem that stumped me...

You have two concentric circles, one with a radius of 1 and one with a radius of 2.

What is the probability that a random chord will pass through the inner circle? Why?

-
How are you defining "random chord?" Do you select two points at random on the outer circle? – Thomas Andrews Apr 8 '11 at 22:26
@Thomas: Could we say we define it such that every possible chord has the same probability of getting chosen? – Mehrdad Apr 8 '11 at 22:27
Nope, that definition doesn't help because the space of possible chords has several different "measures" on it that makes all the chords "seem" equally probable. For example, if you pick a point inside the outer circle, then pick a random direction, you get a different probability measure on the space of chords. – Thomas Andrews Apr 8 '11 at 22:31
@Thomas: Yeah I think that's why the problem tricked me... how about saying, pick two random points in the bigger circle, and choose the chord that goes through them? – Mehrdad Apr 8 '11 at 22:33
For what it's worth, the difficulty of defining that measure yields something called the Bertrand Paradox: en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29 – Thomas Andrews Apr 8 '11 at 22:46

## 4 Answers

This is equivalent to the problem that yields the Betrand Paradox, which shows that how we define a probability measure on a continuous set can greatly affect how we define the probability of an event.

As other people have argued, if you pick two points on the outer circle at random, then the probability that the chord between them hits the inner circle is 1/3.

If you pick two random points inside the outer circle, then the odds that the chord between them intersects the inner circle is at least as great as the probability of one of the two points being inside the inner circle, and that probability is 7/16 = 1-9/16.

-

Consider the general case with the inner circle of radius $r$ and the outer circle of radius $R$, $0<r<R$. Without loss of generality we may assume that one of the points of the cord is the left extreme of the horizontal diameter ( the point with coordinates $(-R,0)$.)

The chord will cut the inner circle if the angle it makes with the horizontal diameter is (in absolute value) less than the angle $\theta$ in the picture. Thus, the probability of cutting the inner circle is $$\frac{2\theta}{\pi}=2\arcsin\frac{r}{R}.$$ When $r=1$ and $R=2$ this gives $1/3$.

This is an example of geometric probability.

Edit

These are two different problems.

1. Find the probability that the chord trhough two randomly chosen points in the larger circle cuts the inner circle.
2. Find the probability that the chord trough two points in the larger disk (the interior of the circle) cuts the inner circle.

My answer is a solution to the first problem.

-
Yup, this is how I did it. Apparently it's not the full story, though! – Mehrdad Apr 8 '11 at 22:53
The chord is defined by one point and he direction. The point ca be arbirarely chosen. Then, if the diecton is chosen with uniform probablity, the result is the one I gave. If instead you consider points in the iner circle, you have to take into account the fact that several points produce the same chord. – Julián Aguirre Apr 9 '11 at 6:40

Another answer: if you define your chord by picking one point in the large circle and making it the center point of the chord, the probability is $\frac{1}{4}$, the ratio of areas of the circles.

-

Hint: The answer is 1/3. Inscribe an equilateral triangle in the larger circle, and the smaller circle just fits. The diagram after equation (2) here helps.

I hope this is not too cryptic!

-
Yup, this is how I did it. Apparently it's not the full story, though! – Mehrdad Apr 8 '11 at 22:53