A pair of points is selected at random inside a unit circle and a line segment is drawn joining the two. Another pair is selected and a second line segment is drawn. Find probability that the two segments don't intersect.

I don't have a clue where to start with this one. I assumed the first pair but I'm not able to come up with the condition for the two segments to not intersect. Please help me out. Thank you.

Here is an example of a possible position of the segments -

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  • $\begingroup$ No idea if it helps, but if you regard the line segments as diagonals of a quadrilateral, then you need the quadrilateral to be convex for the diagonals to intersect $\endgroup$ – Henry Feb 19 '15 at 10:10
  • $\begingroup$ Are the points supposed to land on the circle? $\endgroup$ – drhab Feb 19 '15 at 10:23
  • $\begingroup$ @drhab They may, but they may also land in it. $\endgroup$ – pkwssis Feb 19 '15 at 11:52

The following is an equivalent way of seeing your problem:

Pick randomly any 4 points in a circle, these four points are the vertices of a convex quadrilateral. After that pick randomly two of them and label them as $P,P'$. Label the other two vertices as $Q, Q'$. Then, the segments $PP'$ and $QQ'$ intersect if and only if they are the diagonals of the quadrilateral.

The process I described is equivalent on picking randomly a segment among the six segments whose vertices are the four points. And the probability of taking a diagonal between them is $\frac{2}{6}$.

So the probability that the two do not intersect is $\frac{4}{6}= \frac{2}{3}$.

  • $\begingroup$ My answer was meaningful in the case that the four point make a convex quadrilateral. If the quadrilateral is not convex, then the probability is $0$. So, now one should calculate the probability for the quadrilateral of being convex. $\endgroup$ – Crostul Feb 19 '15 at 12:03

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