Infinite points on a paper? I remember solving questions like this: On a paper with dimensions $30cm$ x $21cm$ if a rubber (erasers)* is dropped, what is the probability that it falls over a grey shaded region of dimensions $1cm$ x $1cm$. 
Now I know that a point is an infinitely small dot, a line segment is essentially a collection of infinite points, and a plane (paper, here) is collection of infinite line segments, and so has infinite number of points too.
Now, the answer to the questions I've solved becomes undefined, i.e. $\frac{\infty}{\infty}$ since there are infinite points on the plane as well as on the shaded region (favourable number of events as well as total number of events become infinity!?)
$^*$Point sized

Can anyone explain me what is wrong here?
Thanks in advance. I know how to solve this method traditionally as Arthur's answer describes. :)

Also, I've recently developed interest in probability, any book referral?
 A: If you pick a random real number, $x$, between $0$ and $1$, what is the probability that $x<\frac{1}{3}$? By your same argument, you'd get the same $\frac{\infty}{\infty}$, yet the result should "obviously" be $\frac{1}{3}$. 
Continuous probabilities are not as simple as dividing the number of "good" outcomes by the total number of outcomes.
You can think of them as limits. For example, you could say that it is not really possible to "pick a random real number from $[0,1]$." But it is possible to pick a random number from $1$ to $2^n$ and divide by $2^n$, resulting in a random variable $x_n$. Then the probability that $x_n<\frac{1}{3}$ is $\frac{\lfloor 2^n/3\rfloor}{2^n}$, which has a limit of $\frac{1}{3}$ as $n\to\infty$.
In the end, if you have a grayed region of area $A$ out of a total paper of area $T$, then the probability that your eraser/rubber will land in your grayed area is $\frac{A}{T}$.
This then becomes a question of "measure theory" (the generalization of the notion of area) and integration.
A: Your sample space $\Omega:=[0,21]\times[0,30]$ contains an infinite number of points $\omega$, and the probability that a particular point, say $(5\pi,\sqrt{137})$, is hit is zero. But the probability $P(A)$ that a given area $A\subset\Omega$ is hit is positive. It is up to you as manufacturer of the model to decide whether $P(A)$ should be proportional to area, i.e.
$$P(A):={{\rm area}(A)\over {\rm area}(\Omega)}$$
(this would be a uniform distribution), or whether you actually aim with your gun at the center of $\Omega$, in which case a so-called probability density $$f:\quad\Omega\to{\mathbb R}_{\geq0},\qquad \int_\Omega f(\omega)\>{\rm d}\omega=1,$$ has to be established (which is large near the center of $\Omega$ and small at the rim). One then has
$$P(A):=\int_A f(\omega)\>{\rm d}\omega\ ,$$
where ${\rm d}\omega$ denotes the area element on your sheet $\Omega$.
A: Divide the paper into a $30\times 21$ grid of $1cm\times 1cm$ squares. Each of the squares has an equal probability that the rubber lands in it, so the probability that it lands in one specific square is therefore $\frac{1}{30\cdot 21} = \frac{1}{630} = 0.15\%$.
This does not answer where you went wrong, only what you should've done to get it right.
