Exercise about the Poisson approximation Suppose that a book of 600 pages contains a total of 240 typographical errors. Develop a Poisson approximation for the probability that three particular successive pages are error-free.
I solved this problem but not sure if the solution is correct.
First let X be the random variable that represents the number of typographical errors in 3 particular successive pages.
Then I have to compute the parameter λ which is (I think is) λ=$\frac{600}{240}=\frac{5}{2}$. 
Therefore Pr(X=0)=$\frac{{(5/2)}^0e^{-5/2}}{0!}=e^{5/2}$.
Could someone check my solution because I'm not sure if it is correct. Thank you  
 A: If $X$ is the number of typos per page, then $E(X) = \lambda_1 = 240/600 = 0.4.$
That is $X \sim \mathsf{Pois}(\lambda_1 = 0.4).$
Solution 1. The probability any one page is free of errors is
$P(X = 0) = e^{-0.4$} = 0.6703$. So the probability three particular
pages are error free is $(0.6703)^3 = 0.3012.$
Solution 2. Let $Y$ be the number of typos per three pages. Then
$Y \sim \mathsf{Pois}(\lambda_3 = 1.2)$ and $P(Y=0) = e^{-1.2} =  0.3012.$
Note: To solve Poisson problems in general, you first figure out the 'domain' of
interest--here three days.  Then find the rate that matches that 'domain'.
So if you wanted the probability that three successive pages have 
two errors or less it would be $$P(Y \le 2) = P(Y = 0)+P(Y=1)+P(Y=2)=0.8795.$$
You can evaluate this by applying the Poisson formula to each term and adding,
or by using software or tables that give Poisson CDFs.  Using R
statistical software, this might look as follows:
ppois(2, 1.2)          # use CDF
## 0.8794871
sum(dpois(0:2, 1.2))   # add three terms of PDF
## 0.8794871

It would be messy to use a method such as Solution 1 to solve this
particular problem. Solution 2 is a more general method.
