Question 1130974
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The probability of <i>exactly</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(k,n,p)\ =\ {{n}\choose{k}}\,\(p\)^k\(1\,-\,p\)^{n-k}]


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


Your number of trials is 12, and the probability of success on any trial is 1 in 4 or 0.25.  Note that "more than 5" is the same as "at least 6", and "less than 4" is the same as "at most 3".  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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