On each page there are $10^4$-many words. Assume the probability to find at most $1$ typo on a page is $\frac{2}{e}$. What is approximately the probability that a word on a page contains a typo?
Let be $S:=\sum\limits_{k=1}^{n}W_k$ a binomially distributed random variable, where $W_k$ shows if a word contains a typo or not. It is $P(S=k)={n\choose k}p^k(1-p)^{n-k}=:b(k;n,p)$. Let be $\pi_{\lambda}(k)$ a Poisson distribution with parameter $\lambda=n\cdot p>0$. We know that the error can be bounded by
$\sum\limits_{k=0}^n|b(k;n,p)-\pi_{\lambda}(k)|\leq 2np^2$.
If we knew that the boundary $2np^2$ would be small then I can use the Poisson distribution to deduce statements about the binomial distribution. If we don't know anything about the quality of the approximation, then how do we show statements like the original question? Is it even possible to answer the question without further assumptions?
Let $X\sim \text{Binomial}(10000,p)$ denote the number of typos on a single page.
You're told that $$\mathbb{P}(X\leq1)=\frac{2}{e}$$
If we approximate $X$ with a $\text{Poisson}(10000p)$ distribution we may say $$e^{-10000p}(1+10000p)=\frac{2}{e}$$
Evidently, the solution to this equation is $p=\frac{1}{10000}$ which is the probability that you seek.