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 $\frac12$?