If and only if you have four points such that no matter which point you choose, it is not in the triangle formed by the other 3, then it is possible to assign the line segments such that they will intersect. Proof of this: Make a triangle, extend the line segments into lines, note that the areas where the fourth point can't be are inside the triangle, and in the areas which meet the triangle at a vertex (As opposed to meeting the triangle at a side). If it's inside the triangle we're done, if it's in one of the areas which ends at a vertex, then that vertex is inside the triangle formed by the other 3.

If A is in the triangle BCD then B is almost surely not in ACD. (And neither is C nor D). In any set of four points, at most one of them can be in the triangle formed by the other 4. Proof of this: Either A or B must be further from line CD, the triangle defined by the closer one clearly cannot contain the further one, since all points in the triangle are closer to CD than the third vertex.

For i=1-4, let x_i be the event "the ith point is in the triangle formed by the other 3". Note that each x_i has probability 11/144. Also, all events are mutually exclusive. Thus, the probability that none of them happen is 1-4*11/144 = 100/144.