This puzzle is an ancient one that doesn't have a definitive answer so far as I can find. Points given to ThunderCloud and plasmid for proofs of possible answers, but will leave the puzzle open for further comments.
I have a criticism of this puzzle, closely related to Pickett's comments, that I haven't seen raised elsewhere: There is no such thing as a random triangle in the plane. How do you pick random points in the plane? We can impose a coordinate system that makes (0, 0) a reference point, and we can add (1, 0) to provides a scale factor and orient the axes, but that's it. What we can't do is pick three arbitrary points in the plane. The origin can be the first point, WOLOG, but the other two, if truly chosen at random, are both points at infinity. It's like asking the average value of the integers. A finite value would, by any measure be disproportionately "close" to the origin. I haven't found any reference to this objection in other discussions of this puzzle. The issue plagues any attempt at a solution.
In the analyses plasmid gave us, the two divergent answers are equally correct -- or equally incorrect. They assume one of the sides of the triangle has finite length. But any random line segment in the plane must have infinite length. And if so, then the analysis compares areas that are both of infinite extent. The analysis ThunderCloud gave us includes the premise that "scale doesn't matter" so let b=1, and then let a and c be anything. Same issue: if b is constrained, we can't let a and c be infinite. Or, if b=1, then it's not random over an infinite space, where things cannot be "scaled".
Other approaches that I found let the triangle, instead, be randomly chosen in a circle. This preserves angular randomness and permits comparison of side lengths, by eliminating the infinity problem. As a bonus, it gives a unique answer.
Follow-on puzzle: What fraction of triangles in a circle are obtuse?