Question 576945
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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\)\left(p\right)^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 5 successes out of 5 trials where the probability of success on any given trial is 15/45 = 1/3, so the odds are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5\left(5,\frac{1}{3}\right)\ =\ \left(5\cr 5\right\)\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0]


But since 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n\cr n\right\)\ =\ 1\ \forall\,n\,\geq\,0\,\in\,\mathbb{Z}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^0\ =\ 1\ \forall\,a\,\in\,\mathbb{R}]


this reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5\left(5,\frac{1}{3}\right)\ =\ \left(\frac{1}{3}\right)^5\ =\ \frac{1}{243}]


Which is the probability, but you asked for the odds.  Odds in favor are the probability of success divided by the probability of failure (which is 1 minus the probability of success), so the odds are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{P_5\left(5,\frac{1}{3}\right)}{1\ -\ P_5\left(5,\frac{1}{3}\right)}\ =\ \frac{\ \frac{1}{243}\ }{\ \frac{242}{243}\ }\ =\ \frac{1}{242}]


1 in 242, or 242 to 1 against.


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