What do I take as the random variable? 
There are 500 misprints in a book of 500 pages. What is the probability
  that a given page will contain at most 3 misprints?

How do I solve this? What do I take as the random variable here?
 A: Lets assume the number of misprints be the
poisson distributed random variable $X$
with $\lambda = 1$.
The probability you are looking for is
$$P(X \leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)= $$ 
$$=\frac{e^{-1}1^0}{0!}
+\frac{e^{-1}1^1}{1!}
+\frac{e^{-1}1^2}{2!}
+\frac{e^{-1}1^3}{3!}$$
A: A brute force approach:
Model the process as balls in urns.  That is, we have $500$ "balls" (=misprints) which we must place in $500$ "urns" (=pages).  Suppose that the placement is wholly independent (which urn one ball goes into has no bearing on any other balls). 
Then...the probability that a given misprint occurs on a given page is $p=\frac 1{500}$ and the probability that a given page holds exactly $k$ misprints is $$Prob(k)=\binom {500}kp^k(1-p)^{500-k}$$  
Using this model, we compute $$Prob(≤3)\sim .98113451$$
The Poisson model gives $\sim .98101184$ so the two methods give extremely similar results.  
To me, the Poisson model is more reflective of what is intended.  That is, I imagine that the probability of misprints is proportional to the number of pages under consideration.  The binomial model ensures that exactly $500$ misprints appear throughout the book which might feel more reflective of the given data...but is it?  When someone says "these $500$ pages contain $500$ misprints" I don't imagine that either count is exact...rather that they are just saying "there's a mean of one misprint per page".  But of course it is a matter of interpretation.  To be sure, the Poisson model is extremely easy to compute with and, as the two models are very close, computational ease is a strong advantage.
To stress: As discussed in the comments below, the ball/urn model is also an approximation (however shoddy the publisher, you really can't have $500$ misprints on one page). In situations like these, it tends to come down to two things:  A. Which model better fits the physical situation (often more an art than a science) and B. computational ease.  
A: Every misprint has a probability of $1/500$ of appearing on the chosen page, and a probability of $499/500$ of appearing elsewhere. The probability $p(k)$ that $k$ misprints appear on that page is then governed by a binomial distribution:
$$
p(k)={500\choose k}\left({1\over500}\right)^k\left({499\over500}\right)^{500-k}.
$$
You then need to compute $p(0)+p(1)+p(2)+p(3)$. 
This case is also well approximated by a Poisson distribution, if that better fits your needs.
A: 
A random variable is a function that assigns a real number to each point in a sample space $S$. It is a numerical-valued variable that represents outcomes in an experiment or random process.

In this case, possible outcomes are the number of misprints on a page, and the sample space is $\{0,1,2, ... ,500\}$ whereby an outcome of $0$ means that there are no misprints on a page, and $500$ means there are $500$ misprints on a page. 
In general, whenever you want to define a random variable, consider the sample space of your experiment and define the random variable such that it assigns a value to each point in the sample space.
Having this sample space of $\{0,1,2, ... ,500\}$, we can define a discrete random variable $X$ as
$$X=number\space of \space misprints \space on \space a \space page$$
Hence, $P(X=x)$ means the probability that there are $x$ number of misprints in a page. Since question is asking for at most 3 misprints, we are looking for $P(X\leq x)$.
To solve this, you can let $X$~$Binomial\big(500,\frac{1}{500}\big)$, whereby $500$ is the number of independent pages, and $\frac{1}{500}$ is the probability of having a misprint.
Thus, 
$$P(X=x)={500\choose x}\bigg(\frac{1}{500}\bigg)^x\bigg(\frac{499}{500}\bigg)^{500-x} $$
Furthermore, since $X$ is a discrete $r.v.$, 
$$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$$
You can approximate this using Poisson, as already discussed on this page, or calculate the sum of probabilities as shown above.
