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This doubt comes from a combinatorics problem in a textbook, which states:

Consider two strictly parallel lines and seven dots, four of which are over one of them, and three over the other. Three dots are chosen at random: what is the probability that they define a circumference?

It turns out, the solution is $1-{{4\choose3}+1\over{7\choose3}}$. That is, all combinations in which the three chosen dots aren't collinear, divided by all combinations of three dots.

My question is: why mustn't they be collinear?

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    $\begingroup$ If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle. $\endgroup$ – Gerry Myerson Mar 20 '19 at 11:45
  • $\begingroup$ isn't a line like a circle with infinite radius $\endgroup$ – Sik Feng Cheong Mar 20 '19 at 11:46
  • $\begingroup$ @SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them. $\endgroup$ – Arthur Mar 20 '19 at 11:49
  • $\begingroup$ The numerator in the "solution," $1-{4\choose3}+1$, is negative. I think you meant $$1-{{4\choose3}+1\over{7\choose3}}$$ $\endgroup$ – Barry Cipra Mar 20 '19 at 12:42
  • $\begingroup$ Have you already seen collinear points on a circle ? $\endgroup$ – Yves Daoust Mar 20 '19 at 14:38
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If three collinear points $ABC$ belong to a circle, then there exists a point $O$ such that $$AO=BO=CO.$$

Let's show this is not possible. From the above equality it follows that $OAB$ and $OAC$ are two isosceles triangles, hence $$ \angle OAB=\angle OBA =\angle OCB. $$ But then, in triangle $OBC$ we have an external angle $\angle OBA$ equal to internal angle $\angle OCB$, and that is impossible by Euclid's exterior angle theorem, QED.

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