Question 1171245
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The probability of *[tex \Large k] successes in *[tex \Large n] independent trials where the probability of success for any single trial is *[tex \Large p] is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(k,n,p)\ =\ {{n}\choose{k}}\,\(p\)^k\(1\,-\,p\)^{n-k}]


The probability of more than *[tex \Large k] successes is then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(> k,n,p)\ =\ \sum_{r=k+1}^n\,{{n}\choose{r}}\,\(p\)^r\(1\,-\,p\)^{n-r}]


Alternatively, it may save some arithmetic to note that *[tex \Large P(> k,n,p)\ =\ 1\ -\ P(<=k,n,p)\ =\ 1\ - \sum_{0}^k\,{{n}\choose{r}}\,\(p\)^r\(1\,-\,p\)^{n-r}]

"At most" means less than or equal which might be easier to calculate using one minus greater than.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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{{n}\choose{r}}
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