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}).

Harrison Mills 2022-09-24 Answered
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 ) .
- Exact Binomial won't work, and the Poisson approximation does not seem appropriate.
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Answers (2)

Zackary Galloway
Answered 2022-09-25 Author has 17 answers
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 Poisson ( λ ) then the distribution of X λ λ approaches the standard normal distribution (i.e. the normal with expected value 0 and standard deviation 1) as λ . In your case, you have λ = 10 4 , which is plenty. If X Poisson ( 10 4 ) then, for example,
Pr ( 9910 X 10050 ) = Pr ( 9910 10000 10000 X 10000 10000 10050 10000 10000 )
Pr ( 9910 10000 10000 Z 10050 10000 10000 ) = Pr ( 0.9 < Z < 0.5 )
where Z N ( 0 , 1 ) .
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videosfapaturqz
Answered 2022-09-26 Author has 3 answers
Step 1
You could use R: for example the probability of being strictly more than 9876 could be about
> p b i n o m ( 9876 , s i z e = 10 11 , p r o b = 10 7 , l o w e r . t a i l = F A L S E ) [ 1 ] 0.8917494
Step 2
This compares with the normal approximation with continuity correction of being above 9876.5 giving the close
> 1 p n o r m ( ( 9876.5 10 11 10 7 ) / 10 11 10 7 ( 1 10 7 ) ) [ 1 ] 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
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