How do you solve a Poisson process problem During an article revision the authors found, in average, 1.6 errors by page. Assuming the errors happen randomly following a Poisson process, what is the probability of finding 5 errors in 3 consecutive pages?
Please explain your methodology, as the main purpose of the question is not getting the answer but the "how to". Please consider this as homework if you will.
Thanks!
 A: If I were his instructor I'd prefer the following explanation, which obviously is equivalent to that given by @cardinal.
Let $N_t$ be the Poisson process of counting the number of errors on consequtive pages, of rate $\lambda=1.6/page$. One supposes that  we observe the process
at integer moments t (here t being assimilated with number of pages). 
Because $P(N_t=k)=e^{-\lambda t}  (\lambda t)^k/k!$, we have $P(N_3=5)=e^{-(3*1.6)}(4.8)^5/5!$.
A: The two most important characteristics of a Poisson process with rate $\lambda$ are


*

*For any interval $(s, t)$, the number of arrivals within the interval follows a Poisson distribution with mean $\lambda (t-s)$.

*The number of arrivals in disjoint intervals are independent of one another. So, if $s_1 < t_1 < s_2 < t_2$, then the number of arrivals in $(s_1, t_1)$ and $(s_2, t_2)$ are independent of one another (and have means of $\lambda (t_1 - s_1)$ and $\lambda (t_2 - s_2)$, respectively).


For this problem let "time" be denoted in "pages". And so the Poisson process has rate $\lambda = 1.6 \text{ errors/page}$. Suppose we are interested in the probability that there are $x$ errors in three (prespecified!) pages. Call the random variable corresponding to the number of errors $X$. Then, $X$ has a Poisson distribution with mean $\lambda_3 = 3 \lambda = 3 \cdot 1.6 = 4.8$. And so
$$
\Pr(\text{$x$ errors in three pages}) = \Pr(X = x) = \frac{e^{-\lambda_3} \lambda_3^x}{x!},
$$
so, for $x = 5$, we get
$$
\Pr(X = 5) = \frac{e^{-\lambda_3} \lambda_3^5}{5!}  = \frac{e^{-4.8} 4.8^5}{5!} \approx 0.175
$$
