You are given two concentric circles $C_1$ and $C_2$ of radius $r$ and $r/2$ respectively. What is the probability that a randomly chosen chord of $C_1$ will intersect $C_2$?

Answer: $1/2, 1/3$ or $1/4$

The first method I used (gives 1/4):

The midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $1/4$

But then I did it by another method and got another answer (1/3):

Choose a point on bigger circle. Now you can get all chords from $0$ to $π$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $π/3$ by some trigonometry. We can do same for every other point so answer is $\frac{π/3}{π}=1/3$

First I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers.

So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points.

Can someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $3$ boundary conditions we have to consider carefully-

  1. If diameters are included or not
  2. Degenerate "chords" that are $r$ distance from center, ie they are actually points on the circumference
  3. If tangents are included in intersection or not

1 Answer 1


I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.

This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.

As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.

Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.

You can get $\frac{1}{2}$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $[0,r]$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $[0,2\pi)$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $<\frac{r}{2}$, which happens exactly $\boxed{\frac{1}{2}}$ of the time.

  • 1
    $\begingroup$ Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord $\endgroup$
    – Amadeus
    Jul 27, 2021 at 6:46
  • $\begingroup$ I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of "things" from the region. $\endgroup$ Jul 27, 2021 at 6:49
  • $\begingroup$ yeah I think a field of mathematics called "measure theory" is used so quite far beyond high school math $\endgroup$
    – Amadeus
    Jul 27, 2021 at 6:54

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