Question 663918
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Just curious:  You said, and I quote: "...to be completely honest."  To me, that means that previous to having made that statement, you were less than completely honest.  So what is it about the earlier part of your post that is dishonest?  Please continue to be completly honest or at least warn me when you begin to dissemble or otherwise communicate falsehoods.


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


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


Where *[tex \LARGE {{n}\choose{k}}] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


The probability of none is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(0,0.27)\ =\ {{15}\choose{0}}\left(0.27\right)^0\left(1\,-\,0.27\right)^{15\,-\,0}]


and the probability of one is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(1,0.27)\ =\ {{15}\choose{1}}\left(0.27\right)^1\left(1\,-\,0.27\right)^{15\,-\,1}]


*[tex \LARGE P(\geq{1})\ =\ 1\ =\ P(0)]


*[tex \LARGE P(\leq{1})\ =\ P(1)\ +\ P(0)]


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