# Binomial probability for large n, small p. I need to compute the probability of getting more than x "successes" in a large number of trials (10^{11}) of an event with a small probability (10^{-7}).

Binomial probability for large n, small p
- I need to compute the probability of getting more than x "successes" in a large number of trials $\left(\phantom{\rule{thinmathspace}{0ex}}{10}^{11}\phantom{\rule{thinmathspace}{0ex}}\right)$ of an event with a small probability $\left(\phantom{\rule{thinmathspace}{0ex}}{10}^{-7}\phantom{\rule{thinmathspace}{0ex}}\right)$.
- Exact Binomial won't work, and the Poisson approximation does not seem appropriate.
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Zackary Galloway
Step 1
I'm guessing that if the Poisson approximation does not seem appropriate, it's because the expected value and the variance (which, for the Poisson distribution, is the same as the expected value) are so big. In that case, approximating it by a normal distribution can serve.
Step 2
If $X\sim \mathrm{Poisson}\left(\lambda \right)$ then the distribution of $\frac{X-\lambda }{\sqrt{\lambda }}$ approaches the standard normal distribution (i.e. the normal with expected value 0 and standard deviation 1) as $\lambda \to \mathrm{\infty }$. In your case, you have $\lambda ={10}^{4},$ which is plenty. If $X\sim \mathrm{Poisson}\left({10}^{4}\right)$ then, for example,
$Pr\left(9910\le X\le 10050\right)=Pr\left(\frac{9910-10000}{\sqrt{10000}}\le \frac{X-10000}{\sqrt{10000}}\le \frac{10050-10000}{\sqrt{10000}}\right)$
$\approx Pr\left(\frac{9910-10000}{\sqrt{10000}}\le Z\le \frac{10050-10000}{\sqrt{10000}}\right)=Pr\left(-0.9
where $Z\sim N\left(0,1\right).$
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videosfapaturqz
Step 1
You could use R: for example the probability of being strictly more than 9876 could be about
$>pbinom\left(9876,size={10}^{11},prob={10}^{-7},lower.tail=FALSE\right)\phantom{\rule{0ex}{0ex}}\left[1\right]0.8917494$
Step 2
This compares with the normal approximation with continuity correction of being above 9876.5 giving the close
$>1-pnorm\left(\left(9876.5-{10}^{11}\cdot {10}^{-7}\right)/\sqrt{{10}^{11}\cdot {10}^{-7}\cdot \left(1-{10}^{-}7\right)\right)}\phantom{\rule{0ex}{0ex}}\left[1\right]0.8915848$
In either case the standard deviation is close to 100 so here only values close to something like $10000±300$ will give interesting probabilities