The first draft of a probability textbook has 625 pages. Assume that number of errors on each page is a Poisson random variable with parameter $\lambda = ln(\frac{5}{4})$ and the number of errors appearing on any given page is independent of the number of errors on any combination of the other pages in the book.
(A) Find the probability that page 254 has at least one typographical error.
(B) Let $T$ be the total number of pages which have at least one typographical error. Give the distribution of the random variable T.
So I solved for (A), and got:
$P(T\geq1)=1-P(X=0) = 1-\frac{(ln(\frac{5}{4}))^{0}e^{-ln(\frac{5}{4})}}{0!}=1-\frac{4}5{}= \frac{1}{5}$
Then for (B), I am confused. I think that $T \sim Binomial(625,\frac{1}{5})$ since $T$ is the total number of pages which have at least one typographical error.
However, I also think that since $T$ is the total number of pages which have at least one typographical error, it also means that $T$ is the total number of pages which have typographical errors.
Then, isn't it $T \sim Poisson[625 \times ln(\frac{5}{4})]$ ?
I don't know which approach is correct. Any suggestions or clarifications is highly appreciated.
$T$ is the total number of pages which have at least one typographical error, so you should think of this as a Binomial random variable, with each page an independent Bernoulli(1/5) random variable.
If you wanted the distribution for the total number of errors, then it would be a Poisson random variable, which is a sum of independent Poisson($\ln(\frac{5}{4})$) random variables, each of which count the number of errors on a given page.