Poisson or Binomial Distribution? A book has $p$ pages and contains $q$ errors. The random variable $X$ is defined as the number of errors in a given page.


*

*What probability distribution law does $X$ follows and why?

*What is its expected value, its variance and its standard deviation?

 A: Sounds like Poisson distribution because it satisfies the properties of a Poisson experiment. I believe mean and variance are the same $\lambda$, and sd is $\sqrt\lambda$.
A: You do have to make an assumption about how errors appear on a page.
If you assume that each of the $q$ errors has an equal probability of appearing on each page and that the page each error appears on is independent of the pages where other errors appear then given $p$ and $q$, for each page the probability distribution is binomial, so $$\Pr(X=x)= {q \choose x}\left(\frac{1}{p}\right)^x \left(1-\frac{1}{p}\right)^{q-x} = {q \choose x}\frac{\left(p-1\right)^{q-x}}{p^q}$$ with mean $\frac{q}{p}$,  variance  $\frac{q(p-1)}{p^2}$ and standard deviation $\sqrt{\frac{q(p-1)}{p^2}}.$
A: Probability distribution function:
$$f_X(n)=\frac{\binom{q+p-1-n}{p-2}}{\binom{q+p-1}{p-1}}$$
for $n \le q$
The formula is equal to the numbers of ways to spread $q-n$ errors over $p-1$ pages divided by the number of ways to spread $q$ errors over $p$ pages.
http://en.wikipedia.org/wiki/Composition_(number_theory)
Expected value is easy:
$$E(X)=\frac{q}{p}$$
Variance and standard deviation will require some calculations...
$$Var(X)=\sum_{n=0}^{q}\frac{\binom{q+p-1-n}{p-2}}{\binom{q+p-1}{p-1}}\cdot n^2-\frac{q^2}{p^2}$$
A: Generally the Poisson Distribution is used to model the number of occurrences of a random variable in a given time interval or in your case the mean number of errors in your book $\ (q/p) $ and in general for events that do not occur very frequently. 
Now a Poisson Distribution is defined to be  
$$ 
\ f(X|\lambda) = \lambda^{x}e^{-\lambda}/x!\,
$$
Where $\lambda$ would be the mean number of errors in your book:
$$
\ \lambda = q/p
$$ 
and your random variable X is then the number of errors on any given page. We can now read the function $f(X|\lambda)$ as the probability of X errors on a page given that the mean number of errors in the book is $\lambda$. 
Finding the expected value of a probability distribution is just a fancy way of asking what is the mean; for the Poisson Distribution that is the same thing as the mean we found earlier. 
The variance can also be found by working out the simple sum
$$ 
E[X(X-1)] = E[X^2]- E[X]  = \sum_0^\infty x(x-1)f(X|\lambda)
$$
*hint along the way do a change of variables $y=x-2$ 
And by definition the variance we conclude that
$$
Var(X) = E[X^2]- (E[X])^2 = \lambda
$$
The standard deviation is taken as the square root of the variance. 
