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 $\frac12$?
 A: I see nothing wrong with the reasoning in Attempt 2, but Attempt 1 is all kinds of wrong.
Just because there are two possible outcomes, it does not follow that the probability of one of them is 0.5. This is only the case when each outcome is as likely as the other, such as with a coin toss.
To randomly pick a third point, out of all the infinite number of points on the plain, that happens to lie exactly on the line AB is hugely unlikely. Infinitely unlikely, in fact. Probability of a triangle = 1.
A: There is no obvious, 'natural' probability distribution of 'choosing points from a plane'. Hence a question starting like

If 3 distinct points are chosen on a plane, what is the probability...

without indicating a specific method of choosing points makes no sense, and the only two answers to it I can think of are 'the probability is any you can think of' or just 'get off'.
A: Nothing can be said about this as long as nothing has been said about the distribution (justifying the comment of angryavian). 
Expressions "at random" or "are chosen" do not speak for themselves because there is no natural uniform distribution on $\mathbb R^2$. 
If the distribution is absolutely continuous wrt the Lebesgue measure (i.e. if the distribution has a PDF) then automatically the answer is $1$ because every line in the plane $\mathbb R^2$ has Lebesgue measure $0$ (which is probably what Kaj means to say). 
So in that case for any fixed line the probability that the third point is chosen on it equals $0$.
A: Attempt 2 is flawed in the assumption that a point C on a line will be between A and B on a line segment.  Since a line stretches on forever in both directions, there are an infinite number of points that could be C that are not between A and B. But this is still not on a plane, just a line.
The actual answer is that the probability of a point being on the line segment that connects any two points "approaches" zero, and because division by infinity is required, IS zero.  In other words, infinitesimally small, and effectively zero.  There are infinite possible points on the line segment, but there are also infinite possible line segments with infinite points in the plane:
$(AB)\infty / (AB)\infty*\infty$
Essentially, you are calculating the odds of a point falling on a specific line segment out of an infinite number of line segments:
$1 / \infty$
But division by infinity is effectively zero.  So, the probability of a third point being on that specific line segment is actually zero.  (This actually makes more sense, when you consider that by definition, a point actually has no size, and a line actually has no width...)
A: 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: This is similar to this probability "joke": 
Given a bowl with 9 black balls and 1 white ball, what's the chance that you pick a white ball? $\frac{1}{2}$, either you pick it or you don't.
While there are indeed both $\infty$ points which are collinear and $\infty$ points which are non-collinear, they're not quite the same $\infty$, so $\frac{\infty}{\infty+\infty}\neq \frac{1}{2}$. (See also Hilbert's Hotel on different levels of infinity)
As a matter of fact, since for collinear points, the choice of $x$ fixes the choice of $y$, there is only one level of infinity. For the non-collinear points, however, there are two levels of infinity: Both $x$ and $y$ can take infinite values. Thus, $\frac{P(\textrm{collinear})}{P(\textrm{non-collinear})}=\frac{1}{\infty}$, which tends to zero. In other words, $P(\textrm{non-collinear})\approx 1$.
A: Having placed the first 2 points, there is a single straight line that goes through these two points. There is also an infinite number of parallel lines that don't go through those 2 points. (The plane itself can be finite).
For a non-triangle to occur, the third point must go on that single line out of the infinite possibilities. Assuming independence the probability of this is zero. Hence the probability of a triangle is 1. 
A: What has happened to good old probability without measure theory ? You can't say that the question is dumb just because a less-than-100-year-old theory can't solve it. If a theory doesn't answer a question, then don't try to use it. The question is actually very good, if you ask me. No need to go into debate about different kinds of infinity. If this is a school-level problem, then it's assumed that each point on the plane is equally likely. If highly skilled  mathematicians can't solve the problem without telling this young fellow about Lebesgue measure or absolute continuity and whatnot then just use conditional probability.
We don't care where the first two points $A$ and $B$ end up. So assuming there are two points on the plane, what is the probablity that the third one $C$ form a triangle ? All that matters is, of all lines parallel to $(AB)$, which one is $C$ on ? If all points are equally likely, then so are all those lines. Looking at all those parallel lines is like looking at $\mathbb{R}$. And $(AB)$ is one point of the real-line. So the problem is like saying "Given $x$ in $\mathbb{R}$ what is the probability for a random $y$ that $y=x$ ?" If you think it's $0$ and you need a measure to prove it, just say :
For any bounded interval $I$ containing $x$ probability that $y=x$ given that $y\in I$ is zero with respect to normalised Lebesgue measure on that interval. So without even assuming that $y\in I$, not a chance.
A: Let's make it simple. No need to bother with any calculation here. Probability is equal to 1 by definition.
Indeed, whenever you have 3 distinct points in a plane, you have a triangle.
And even in the case where the 3 points are aligned on the same line, we are just facing with a degenerate triangle.
P.S: this also applies in a 3D space.
A: 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. 
A: Eventhough, the actual answer is 1, because the measure of any line is 0, I am wondering whether the actual formulation of the problem was just slightly different since it looks silly the provided textbook answers to be 1/2.
One suggestion for the true formulation is the following:
Given 3 points A,B and C, taken at random in the plane(here, to make the things proper, one may substitute that with 'uniformly at random in a sqaure $C_{n}$ with side n, $n\to \infty$'), what is the probability that ABC is a positive oriented triangle?
The variants where just one word as 'obtuse' or 'acute' in front of 'triangle' is missed doesn't make sense since in such cases the answer differs from 1/2.
