How to solve this probability distribution?

Question

A document containing 160 pages has 200 errors. Find the probability that on page 102 and 103 will contain exactly three errors.

Is it using binomial distribution? I have assigned p = 3/200 and q = 197/200

Is it correct ?

Please help me

Answer 1

There are $200$ errors and each of them "chooses" a page to stand on.

For every error the $160$ pages have equal chances to be chosen, so that the probability that it becomes one of the pages $102,103$ (a succes) equals $\frac2{160}=\frac1{80}$.

To be found is now $\Pr(X=3)$ where $X$ denotes the number of errors that chooses one of the pages $102,103$.

The distribution is binomial with parameters $n=200$ and $p=\frac1{80}$.

Answer 2

As you try to solve this problem you soon realize that you must make some assumptions in order for it to be tractable. You tagged your question with two distribution tags. I gave up with one, so perhaps we can make hay with the Poisson distribution model.

OK here goes. You have 200 errors per 160 pages, or, on average, 200/80 = 2.5 errors per 2 pages. This feel promising since the question was asking for the probability of exactly 3 error per a given two page look-see.

Using the discrete Poisson model,

$ P(k{\text{ errors on two pages}})=e^{-\lambda }{\frac {\lambda ^{k}}{k!}} $, where $\lambda = 2.5$

So plug in $3$ for your answer.

Great, we got an answer. Now you have to justify why using this distribution makes sense.

The question starts off saying the document HAS 200 errors. What would make more sense is if we said that past experience with our typist leads us to estimate $\lambda$ as follows,...blah...blah...Poisson...blah...blah...blah...