How Many Searchers to Hire

Key idea. Keep hiring while one more finder adds more expected recovery than the $5000 he costs.

If $n$ finders are hired, the drive is lost only if every one of them fails. They act independently, so this has probability $(0.1)^n$ . The expected profit is therefore

E(n) = 500000\bigl(1-(0.1)^n\bigr) - 5000n .

The $k$ -th finder only matters when the first $k-1$ have all failed. This happens with probability $(0.1)^{k-1}$ , and he then succeeds with probability $0.9$ . His marginal expected gain is

E(k)-E(k-1) = 500000\cdot 0.9\cdot(0.1)^{k-1} - 5000 = 450000\cdot 10^{-(k-1)} - 5000 .

This is strictly decreasing in $k$ , so $E$ is unimodal. The first three values are:

$k=1$ : $450000-5000>0$ ;
$k=2$ : $45000-5000>0$ ;
$k=3$ : $4500-5000<0$ .

Every later increment is also negative. So $E(n)$ rises up to $n=2$ and falls afterwards. As a check:

$E(1)=445000$ ;
$E(2)=495000-10000=485000$ ;
$E(3)=499500-15000=484500$ .

\boxed{n=2,\ \text{with expected profit } \$485000}