Question 1007598
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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)!}]


(a) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_8(0,0.6)\ =\ {{8}\choose{0}}\left(0.6\right)^0\left(1\,-\,0.6\right)^{8\,-\,0}]


(b) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_8(8,0.6)\ =\ {{8}\choose{8}}\left(0.6\right)^8\left(1\,-\,0.6\right)^{8\,-\,8}]


(c) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_8(>1,0.6)\ =\ 1\ -\ \sum_{k=0}^1\,{{8}\choose{k}}\left(0.6\right)^k\left(1\,-\,0.6\right)^{8\,-\,k}]


(d) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_8(\geq 6,0.6)\ =\ \sum_{k=6}^8\,{{8}\choose{k}}\left(0.6\right)^k\left(1\,-\,0.6\right)^{8\,-\,k}]


You can do your own arithmetic.  Hints:  *[tex \Large {{n}\choose{0}}\ =\ {{n}\choose{n}}\ =\ 1]. And *[tex \Large x^0\ =\ 1\ \forall x\ \not =\ 0\ \in \mathbb{R}] 


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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