Question 875115
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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)!}]


You need to calculate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5\left(5,\frac{1}{3}\right)\ =\ {{5}\choose{5}}\left(\frac{1}{3}\right)^5\left(1\,-\,\frac{1}{3}\right)^{5\,-\,5}]


As an aid to doing this calculation, remember that *[tex \LARGE {{n}\choose{n}}\ =\ 1] and anything raised to the zero power is 1.


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