If we select three independent and random points $A$,$B$ and $C$ in a plane, what shall be the probability that they are collinear?
Actually, this problem was asked to my friend in an interview, he applied common sense and approached in the following way : since any two, say $A$ and $B$ lie on a line, $C$ either lies on the line joining $A$ and $B$ or anywhere else in the plane.Now since the plane is so vast compared to the line (here comes the argument) the probability must be tending to 0.
But the Mathematics professor (interviewing him) said that as both, the line and the plane stretch to infinity we cannot compare them.Therefore there are only two possibilities – $C$ will either lie on the line or on the plane.So, probability is equally distributed.Thus answer will be 0.5.
The whole argument lies within the point Whether two infinities can be compared or not.
Although he also wasn't satisfied but he couldn't argue further because he didn't knew two Infinities can be compared!(For example the set of all Real Numbers is strictly bigger than set of all Natural Numbers which can be proved by Cantor's Diagonal Argument)
We can compare cardinality of two infinities and tell which one is relatively bigger and I couldn't find any One-One and On-to function for the points of line and the plane and I am pretty sure I wouldn't find any.(To explain that they are equal)
But as the interviewer himself was a Mathematics professor, I couldn't simply neglect his answer.
EDIT : I previously had accepted the answer given by sds , but recently I came to know that, the cardinality of the set containing all the points of the plane, and the set containing all the points on the line is, infact equal, and hence we cann't say that plane is "vast" as compare to line. So, I am again confused.
Best Answer
Probability
Neither Math Professors nor information transmission are infallible. What you are ascribing to a Math Professor makes little sense. Your friend is right, both in the answer (the probability is $0$) and in his argument (dimension).
Specifically, a triple of points in a plane forms a 6-dimensional space, and the subspace of collinear points is 5-dimensional, so its measure (probability) will be $0$ for any "reasonable" definition of probability.
"The number of points" in both 6d space and 5d subspace does not enter the picture (the cardinality of both is continuum and is irrelevant).
Strictly speaking, it is not easy to define a probability measure on the whole plane. The measure cannot be both invariant under euclidean transformations and have the total probability of $1$. Thus the usual solution is to restrict the points to a bounded subset.
Thus, let the points be selected uniformly and independently from the unit square $[0;1]\times[0;1]$ with the usual Lebesgue measure (aka area) as the probability.
Then the 3 points $A=(x_1,y_1), B=(x_2,y_2), C=(x_3,y_3)$ can be represented as a single point selected uniformly from $[0;1]^6$.
The subset of collinear points is described by a single equation
$$ \det \begin{bmatrix} x_1-x_2 & x_1-x_3 \\ y_1-y_2 & y_1-y_3 \end{bmatrix} = 0 $$
which defines a co-dimension 1 subset and thus has probability $0$.
Comparing infinities
Your are asking, in bold, whether two infinities can be compared or not.
The answer is definitely YES.
It depends, however, on what you are trying to accomplish.
E.g., both integers and reals are infinite sets, but integers are countable while reals are not. Thus one can say that there are more reals than integers. (Moreover, this idea of cardinality gives us a very simple argument why irrationals and transcendentals exist: because the "alternatives" (rational and algebraic numbers) are countable and reals are not).
However, you are talking about probabilities which means counting only for finite sets. For infinite sets you have to use measure theory.
Here you would say that when selecting a point at random from the segment $[0;1]$, the probability that you will end up with a number between, say, $\frac{1}{3}$ and $\frac{1}{2}$ is the length of the segment $[\frac{1}{3};\frac{1}{2}]$ which is $\frac{1}{6}$. Here we just compared two (uncountable) infinities: the set of points in $[0;1]$ and the set of points in $[\frac{1}{3};\frac{1}{2}]$ (both continuums).