A mathematics textbook has 100 pages on which typographical errors in the equations could occur. Suppose there are in fact two pages with errors. What is the probability that a random sample of 20 pages will contain at least one error?
read some solutions online, it says it is poisson, why is it? and i seem don't quit understand enter image description here
\begin{align} P(\text{at least one error})&=1-P(\text{no error})\\ &=1-\dfrac{\binom20\binom{98}{20}}{\binom{100}{20}} \end{align} i.e. choose $0$ out of $2$ pages with errors and choose $20$ out of $98$ correct pages.
However, since the number of wrong pages is small compared to the total number of pages, Poisson can be used as a quick approximation. (But for this case, it's not very appropriate anyway.)
This reminds me of the same birthday problem.
$1 - (\frac{98}{100}\cdot \frac{97}{99}..............\frac{79}{81}) = 1 - (\frac{98!}{78!}\cdot \frac{80!}{100!}) = .3612$
