Just a few chords, please...

This is just a quick outline of one possible solution method.

For any $6$ distinct points on the circumference of a circle we can form $15$ distinct sets of 3 chords where each of the $6$ points is the endpoint of precisely one of the chords. (We don't really need to worry about the cases where we can form 3 chords using only $4$ or $5$ distinct points on the circle as these will contribute nothing from a probabilistic standpoint.)

Now with only $15$ distinct sets to consider we can just carefully draw out all the different scenarios, (avoiding having 3 chords intersect at the same point), and make a tally of the number of points of intersection in each case. Doing this we find that

$p(0) = \frac{5}{15}, p(1) = \frac{6}{15}, p(2) = \frac{3}{15}, p(3) = \frac{1}{15}$ ,

and thus $p(0) - p(1) + p(2) - p(3) = \frac{1}{15}$ .

This gives us $a = 1, b = 15$ , and thus $a + b = \boxed{16}$ .

(Ideally I would provide illustrations of the $15$ cases under consideration but this is beyond my technical capabilities at the moment, so apologies for leaving you on your own in that regard.)

EDIT: I'll try my best to describe the cases without the aid of illustrations. First, draw six points "randomly" on the circumference of a circle and number them 1 to 6 clockwise. (By "randomly" I just mean that the spacing between points should be random, as regular spacing would result in cases where 3 chords intersect at 1 point, which we would prefer to avoid.) Starting with point 1, we can join it to any of the other 5 points to form our first chord. Once this is done, we take the next unconnected point (going clockwise) and join it to one of the other 3 remaining unconnected points to form our second chord. Finally, join the last 2 remaining unconnected points to form our third chord. Using this process, we end up with a total of $5 * 3 = 15$ possible distinct sets of 3 chords formed using the 6 points. (If we were to look at the same question with 4 chords we would start with 8 endpoints, resulting in $7 * 5 * 3 = 105$ possible sets of 4 chords. For 5 chords there would $9 * 7 * 5 * 3 = 945$ possible sets of 5 chords. This would be too many to draw by hand, so we would probably need to write a computer program to help us when the number of chords exceeds 3.)

Now we just count the total number of points where the chords intersect inside the circle. The 5 sets that result in 0 points of intersection are as follows:

$[(1,2),(3,4),(5,6)], [(1,6),(2,3),(4,5)], [(1,2),(3,6),(4,5)], [(1,6),(2,5),(3,4)], [(1,4),(2,3),(5,6)]$ .

These 5 sets out of the 15 possible sets indicate that $p(0) = \frac{5}{15}$ .

Two points of intersection are obtained with the following sets:

$[(1,3),(2,5),(4,6)], [(1,4),(2,6),(3,5)], [(1,5),(2,4),(3,6)]$ .

Thus $p(2) = \frac{3}{15}$ . To obtain 3 distinct points of intersection, we only have one set that does the job, namely $[(1,4),(2,5),(3,6)]$ . This means that $p(3) = \frac{1}{15}$ .

Now since we must have $p(0) + p(1) + p(2) + p(3) = 1$ , we must have $p(1) = \frac{6}{15}$ . I will leave it to the reader as an exercise to establish the 6 sets of chords that each result in 1 point of intersection, just to verify that we haven't miscounted anywhere.

Here is a link that provides more information on this problem for the general case of $n$ chords, complete with a (complicated) closed-form formula: link