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Four points are chosen independently and at random on a circle. Find the probability that chords X1X2 and X3X4 intersect a) without calculation using a symmetry argument and b) from the definition by an integral

I'm lost here. I'm thinking of using some kind of angle argument, but not sure how to go about it.

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@user6312 - I think the definition of a "chord" is the part of the line inside the circle, so that's redundant. –  Thomas Andrews Apr 18 '11 at 22:00

2 Answers 2

Without calculation, you can see that given any four points on the circle, there are three ways of partitioning them into two pairs of points, and one way yields an intersection between the chords, the other two yield a non-intersection between the chords. So the probability is $\frac{1}{3}$.

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If $\alpha\in[0,2\pi]$ is the angle from $X_1$ to $X_2$, then the probability that $X_3$ is between $X_1$ and $X_2$, and $X_4$ is between $X_2$ and $X_1$, which is one of the configurations were the chords intersect, will be $\dfrac{\alpha}{2\pi} \cdot \dfrac{2\pi-\alpha}{2\pi}$.

Averaging this for all (equiprobable) values of $\alpha$, we find $$\dfrac{\int_0^{2\pi} \dfrac{\alpha}{2\pi} \cdot \dfrac{2\pi-\alpha}{2\pi} d\alpha}{2\pi} = \left.\left(\dfrac{2\pi\dfrac{\alpha^2}2-\dfrac{\alpha^3}3}{(2\pi)^3}\right)\right|_0^{2\pi}=\dfrac16$$

The probability that $X_4$ is between $X_1$ and $X_2$, and $X_3$ is between $X_2$ and $X_1$ will also be $\dfrac16$, so the probability that there is an interesection will be $2\times\dfrac16=\dfrac13$, consistent with the intuitive answer given by a symmetry argument.

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