3.4.2 problem 47

(a)

Consider the simple case of $m<\frac{n}{2}$. Then, the trays don’t have enough pages to print $n$ copies. Desired probability is $0$.

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,

$$P=\{\begin{array}{cc}0& m<\frac{n}{2}\\ \mathrm{pbinom}(m,n,p)-\mathrm{pbinom}(n-m-1,n,p)& \frac{n}{2}\le m<n\\ 1& m\ge n\end{array}$$

(b)

Typing out the hinted program in the R language, we get that the smallest number of papers in each tray needed to have $95$ percent confidence that there will be enough papers to make $100$ copies is $60$.