random triangle

Picking points "at random" inside a triangle is a bit tricky, and is discussed at length here .

Later on in this article it is mentioned that the expected area of the convex hull of a polygon formed by $3$ points, (i.e., a triangle), chosen independently and uniformly inside a triangle of unit area is $\dfrac{1}{12}$ . Now since we have $4$ points to choose, we have $4$ independent scenarios where one of the chosen points lies within the triangle formed by the other $3$ points. Thus the probability that one of scenarios plays out is $4 * \dfrac{1}{12} = \dfrac{1}{3}$ .

Thus $a = 1, b = 3$ and $a + b = \boxed{4}$ .