Exercise 3.47

On the other hand, if $m\ge n$, then desired probability is $1$, since each tray individually has enough pages.

Now, consider the more interesting case that $\frac{n}{2}\le m<n$. Associate $n$ pages being taken from the trays with $n$ independent Bernoulli trials. Sample from the first tray on success, and sample from the second tray on failure. Thus, the assignment of trays can be modeled as a Binomial random variable, $X\sim$ Bin$(n,p)$. As long as not too few pages are sampled from the first tray, the remaining pages can be sampled from the second tray. What is too few? $n-m-1$ is too few, because $n-m-1+m<n$.

Hence,