Question 1158395
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The probability of <b>exactly</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)\ =\ {{n}\choose{k}}\,(p)^k\,(1-p)^{n-k}]


The probability of <b>at least</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\(\geq{k},n,p\)\ =\ \sum_{r=k}^n{{n}\choose{r}}\,(p)^r\,(1-p)^{n-r}]


The probability of <b>at most</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\(\leq{k},n,p\)\ =\ \sum_{r=0}^k{{n}\choose{r}}\,(p)^r\,(1-p)^{n-r}]


For your problem, *[tex \Large n\ =\ 13], *[tex \Large k\ =\ 10], and *[tex \Large p\ =\ 0.9]


You can do your own arithmetic.
								
								
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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