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?

 A: 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.
A: 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.


*

*Find the probability that the chord trhough two randomly chosen points in the larger circle cuts the inner circle.

*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.
A: 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.
A: 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!
