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I know One can calculate the probability of getting at least $k$ successes in $n$ tries by summation: $$\sum_{i=k}^{n} {n \choose i}p^i(1-p)^{n-i}$$ However, is there a known way to calculate such without summation?

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Yes, Brian, thank You. I will edit. – xuinkrbin. Mar 13 '13 at 15:07
up vote 3 down vote accepted

To my knowledge, there is no easier way for an exact evaluation of this formula.

However, when $n$ is large, the binomial distribution is close to a normal distribution, which can be used for an approximation.

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Though one never has to sum more than $\lceil n/2\rceil$ terms, since one can take complements. – Brian M. Scott Mar 13 '13 at 15:05
@azimut: Then, for sufficiently large n, I could take the integral from k to n, yes? – xuinkrbin. Mar 13 '13 at 15:16
@Brian M. Scott: Unfortunately, the n which I am considering could run anywhere from 12 to positive infinity. :-/ – xuinkrbin. Mar 13 '13 at 15:17
@xuinkrbin: I suspected something like that; I just thought that it ought to be mentioned. You might want to look at this. – Brian M. Scott Mar 13 '13 at 15:21
@xuinkrbin: I’d expect so, though I don’t know whether that really gains you much. (This is getting pretty far outside the areas in which I’m comfortable.) – Brian M. Scott Mar 18 '13 at 20:54

If you are doing an "at least" from 12 to positive infinity, you could also do 1-p( x < 11), no?

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No, the [12, +infinity) is for n; not k. – xuinkrbin. Mar 14 '13 at 0:18

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