Approximating $1-(1-P)^N$ for $PN\ll1$ I have a Bernoulli trial where $P$ is really small $\approx 10^{-20}$ and $N\approx10^{10}$ is really big but still much smaller than $1/P$, ie $NP$ is very small.
Directly computing the chance of at least one success, $1-(1-P)^N$, is not practical (AFAIK).  My intuition says its very close to $PN$ though (every time you multiply $(1-P)$ times something close to $1$, the product is $P$ smaller).  Is this sufficient?  Is there something I can cite to say this is sufficient or is it “obvious”.
EDIT: Addressing a comment below.  By “very close” I mean within an order of magnitude.
 A: By binomial approximation, $(1-P)^N\approx 1-PN$ when $|P|<1$ and $|PN|<<1$ .
Hence, $1-(1-P)^N\approx 1-(1-PN)=PN$
If we want more accuracy, we can use binomial expansion to get a more accurate approximation. For example
$$(1-P)^N\approx 1-PN+\frac{N(N-1)}{2}P^2$$
$$1-(1-P)^N\approx PN-\frac{N(N-1)}{2}P^2$$
However, for most purposes the first approximation is sufficient and since it is algebraically simple it can sometimes lead to nice simplifications when dealing with such scenarios.
A: A nice approach is to just give a bracket that is as narrow as you want. One way to do that is to let $f(P)=(1-P)^N$ and then $f'(P)=-N(1-P)^{N-1},f''(P)=N(N-1)(1-P)^{N-2}$, so $f(0)=1,f'(0)=-N,f''(0)=N(N-1)$. Additionally $f'''(P)<0$. Therefore, the first order Taylor expansion is a lower bound (since the second derivative is everywhere positive) and the second order Taylor expansion is an upper bound (since the third derivative is everywhere negative). Hence
$$1-NP \leq (1-P)^N \leq 1-NP+N(N-1) P^2/2$$
or to specialize to your case
$$NP-N(N-1) P^2/2 \leq 1-(1-P)^N \leq NP.$$
Therefore the relative error in choosing an approximation from this interval is at worst on the order of $NP$.
A: Let $f(P):=(1-P)^N$ for $N\ge1$ and $0\le P\le 1$. We have $f'(P)=-N(1-P)^{N-1}$, $f''(P)=N(N-1)(1-P)^{N-2}$ and by Taylor's theorem,
$$\left\lvert PN-\left(1-(1-P)^N\right)\right\rvert=\left\lvert f(P)-f(0)-Pf'(0)\right\rvert\le\frac{P^2}2\sup_{[0,P]}\,\lvert f''\rvert=\frac{P^2}2N(N-1).$$
So the error you make when approximating $1-(1-P)^N$ by $PN$ is of order $P^2N^2$.
A: As other answerers, using Taylor series, we have
$$1-(1-P)^N=PN \Bigg[1-\frac{1}{2} (N-1) P+\frac{1}{6} (N-2) (N-1) P^2+O\left(P^3\right) \Bigg]$$
Now, we can transform the quantity in brackets as a $[n,n]$ Padé approximant. The simplest would then give
$$1-(1-P)^N\sim P\,N\,\, \frac{1-\frac{1}{6} (N+1) P}{1+\frac{1}{3} (N-2) P} $$
