Question 1127552
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You need the probability of seven successes out of 22 trials, plus eight successes out of 22 trials, plus nine...and so on up to 22 successes out of 22 trials.


The probability of *[tex \Large k] successes out of *[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}]


So you need the sum:


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


For at least 7 out of 22, you will have a sum with 16 terms.  The following sum, which gives you the same result only requires calculating 8 terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  P\(\geq{k},n,p\)\ =\ 1\ -\ P\(<k,n,p\)\ =\ 1\ - \sum_{r=0}^k\,{{n}\choose{r}}\,\(p\)^r\(1\,-\,p\)^{n\,-\,r}]


Make sure your calculator has fresh batteries.
								
								
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">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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