Finding n for a given P of a Bernoulli trial

I'm randomly sampling $N$ items and I want to find $n$ such that I have a probability $P$ that I'll miss one. Practically, I'd select $P$ to be something like $10^{-12}$ so I'm almost assured to sample everything from among a relatively small sample set of $n$ things (this is for some program tester).

Anyways, using the Bernoulli trials equation of

$$P = \binom{n}{k}p^kq^{n-k}$$

where $k$ = 1, and $p = N^{-1}$, I think I've simplified it correctly as shown:

\begin{align}P &= \frac{n!}{(n-1)!}\left(\frac1N\right)\left(1-\frac1N\right)^{n-1} \\ P &= \frac{n}{N}\left(\frac{N-1}{N}\right)^{n-1} \\ PN^n &= n(N-1)^{n-1} \\ n \ln (PN) &= \ln(n) + (n-1)\ln(N-1) \end{align}

Now I don't know where to go; I can't seem to solve for $n$. The second line seems closer, but I can't seem to figure out how to solve for $x$ given something like $xk^x$ (I could also get this I think:)

$$\left(N-1\right)P = n\left(\frac{N-1}{N}\right)^n$$

• Do you want to 'miss $exactly$ a $particular$ one' , miss exactly one, or miss at least one (that is, not get them all)? How do you get from your first to your second equation? Commented Aug 31, 2015 at 21:12
• I want the probability that I will miss exactly one (then as $n$ continues to rise, $P$ will continue to fall). If $k=1, p=N^{-1}$ and $q = 1-N^{-1}$, doesn't that go from 1 to 2? Commented Aug 31, 2015 at 21:56
• Maybe express $n = aN$. For large $n,$ one has $(1 - a/n)^n \approx e^{-a}.$ Commented Aug 31, 2015 at 23:24