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Probability problem

I've just realised I completely misunderstood the meaning of this week's "Touch me not" problem (Geometry and Combinatorics). Instead of finding the probability that the three sectors have a point in common (other than the centre), I tried to find the probability that at least two sectors have a point in common. I'd like to propose this modified version of the problem for discussion to see how other people would approach it. (I couldn't find a nice elementary solution: I had to compute some definite integrals)

I'll report the modified text here: "Three sectors are chosen at random from circle \(C\), having angles \(\frac{\pi}{10},\frac{2 \pi}{10}, \frac{3\pi}{10}\) respectively. What is the probability that at least two of these three sectors have a point in common other than the center of the circle.

Note by Lorenzo Sarnataro
4 years, 3 months ago

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In either version of the problem, using indefinite integral makes the calculation pretty nasty.

Check out Geometric Probability to transform the question into one about volume (of if you're good, area).

The difficulty associated with geometric probability usually comes from one of two areas, the first is finding a good way to model the problem geometrically, and the second is in trying to determine the areas/volumes of particular regions in order to calculate the relative probabilities. As in finite probability, it is sometimes simpler to find the probability of the complement.

Calvin Lin Staff - 4 years, 3 months ago

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