Probability that 3 points in a plane form a triangle This question was asked in a test and I got it right. The answer key gives $\frac12$.


  Problem: If 3 distinct points are chosen on a plane, find the probability that they form a triangle.


Attempt 1: The 3rd point will either be collinear or non-collinear with the other 2 points. Hence the probability is $\frac12$, assuming that collinearity and non-collinearity of the 3 points are equally likely events.

Attempt 2: Now suppose we take the midpoint (say $M$) of 2 of the points (say $A$ and $B$). We can draw an infinite number of lines passing through $M$, out of which only 1 line will pass through $A$ and $B$. Keeping this in mind, we can choose the 3rd point $C$ on any of those infinite lines, excluding the one passing through $A$ and $B$. Now it seems as if the probability will be tending to 1.

What is wrong with attempt 2? Or is the answer actually 1 and not $\frac12$?
There is no such thing as a uniform distribution on the plane.  Without specifying how the points are chosen, the question is not properly stated.  However, if the points are chosen independently from some continuous distribution (absolutely continuous with respect to Lebesgue measure), the probability of the third point lying exactly on the line through the first two is $0$.
A possibly silly way to finesse the ambiguity in the initial distribution: put the points on the projective plane. Then uniform distribution is well-defined and clearly, in the presence of uniform distribution, it is legitimate to use it to define a set of three points chosen randomly. This, of course, again leads to the probability of three points forming a triangle being 1.