Question 388086
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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)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE \left(n\cr k\right\)] 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 want *[tex \Large n\ =\ 5], *[tex \Large k\ =\ 0], and *[tex \Large p\ =\ \frac{1}{6}], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5\left(0,\frac{1}{6}\right)\ =\ \left(5\cr 0\right\)\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{5}]


Note that *[tex \LARGE \left(n\cr 0\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{N}] and *[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}] 


So the calculation reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5\left(0,\frac{1}{6}\right)\ =\ \left(\frac{5}{6}\right)^{5}]


A little calculator work and you are there.


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