Question: Let a book of 500 pages contain 30 printing mistakes. Let
these errors are randomly distributed throughout the book and the
number of errors per page has a poisson distribution. The probability
that 10 pages selected at random is free from error is ______________.
($1^{st}$ approach):
Error probability(error per page) $p = \frac{30}{500} = 0.06$
No. of pages to consider $n = 10$
So, poisson parameter $\lambda = n * p = 0.6$
Now, as distribution is poisson so, $P_X(x) = \frac{e^{-\lambda}*\lambda^x}{x!}$
So according to that required probability = $P_X(0) = e^{-0.6} = 0.5488$ .
$2^{nd}$ approach:
Now, let's consider probability of no error in page. So $p = 0.94$
And we have to consider $10$ pages so $n = 10$.
Now, $\lambda = n*p = 9.4$
Now required probability is(same as all 10 pages have no errors) $P_X(10) = \frac{e^{-9.4}*9.4^{10}}{10!} = 0.1227$
Please point out the reason of this mismatch. Thanks.
Best Answer
General statement: both approaches are wrong, although the first is a good approximation, using Poisson distribution for all the errors (per page is irrelevant). The second approach makes no sense to me.
Approximate answer: prob. no errors on one page $=.94$, so probability no errors on ten pages $=.94^{10}=0.5386$ This is approximate since independence is assumed for number of errors per page.
Exact answer (no independence assumption): prob. of no errors on ten pages $=(1-\frac{30}{500})(1-\frac{30}{499})...(1-\frac{30}{491})$ Here the basic idea is that if there are no errors on a given page, the probability of an error on another page is slightly increased.