[Math] Exercise about the Poisson approximation

poisson distributionpoisson processproof-verificationstatisticsstochastic-processes

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

Best Answer

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.

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