I almost made this problem too difficult. How are the points chosen? Random xcoordinate from sqrt(20) to sqrt(20) and then random +/y? Random angle from the origin? It doesn't matter.
The relative positions are all that matters.
(i) is clearly zero.
(ii) requires the order of the points not be pairwise (AB)(CD)(EF)
Lets consider picking the points and then applying labels in such a way that no pairs meet.
There are 6!/2/2/2=90 ways to do this.
If we call one A then there are 2 choices for B.
We have 4 choices for C but this fixes D.
Finally 2 choices for E fixes F.
2*4*2=16 so there's a 16/90=8/45 chance that two two meet and a 37/45 chance that at least two meet.

Posted by Jer
on 20160609 14:19:55 