Question 358945
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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)!}]


The probability of at least one is the probability of exactly one plus the probability of exactly two plus the probability of exactly three plus the probability of exactly four.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4\left(\geq1,\frac{18}{39}\right)\ =\ \sum_{i=0}^4 \left(4\cr i\right\)\left(\frac{18}{39}\right)^i\left(\frac{21}{39}\right)^{4\,-\,i}]


But since that sum plus the probability of exactly zero occurrences equals one, we can calculate the sum by subtracting the probability of exactly zero from 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4\left(\geq1,\frac{18}{39}\right)\ =\ 1\ -\ P_4\left(0,\frac{18}{39}\right)\ =\ 1\ -\ \left(4\cr 0\right\)\left(\frac{18}{39}\right)^0\left(\frac{21}{39}\right)^{4}]


But since *[tex \LARGE \left(n\cr 0\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{N}] and *[tex \LARGE x^0\ =\ 1\ \forall x\ \in\ \mathbb{R}], the above reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 1\ -\ \left(\frac{21}{39}\right)^{4}]


You can do the rest of the arithmetic.


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