Question 1166409
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given 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 <b><i>at most</b></i> *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p] is given by:


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


The probability of <b><i>at least</b></i> *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p] is given by:


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


The probability of <b><i>more than</i></b> *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p] is given by:


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


Make adjustments to the summation indices depending on exactly what you are looking for.  Also note the following relationship that may save you some arithmetic:



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

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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