Question 1028846
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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)!}]


So the probability of 3 Yes results would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3(3,0.2)\ =\ {{3}\choose{3}}\left(0.2\right)^3\left(0.8\right)^{0}\ =\ 1\ *\ 0.008\ *\ 1\ =\ 0.008]


The probability of 2 Yes results (which could be YYN, YNY, or NYY), would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3(2,0.2)\ =\ {{3}\choose{2}}\left(0.2\right)^2\left(0.8\right)^{1}\ =\ 3\ *\ 0.04\ *\ 0.8\ =\ 0.096]


Each of YYN, YNY, or NYY would be 1/3 of the above result.


The probability of 1 Yes result (YNN, NYN, or NNY) would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3(1,0.2)\ =\ {{3}\choose{1}}\left(0.2\right)^1\left(0.8\right)^{2}\ =\ 3\ *\ 0.2\ *\ 0.64\ =\ 0.384]


Each of YNN, NYN, and NNY would be 1/3 of the above, so P(NNY) = 0.128


The probability of Zero Yes results, NNN, would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3(0,0.2)\ =\ {{3}\choose{0}}\left(0.2\right)^0\left(0.8\right)^{3}\ =\ 1\ *\ 1\ *\ 0.8\ =\ 0.512]


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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