Question 312098
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You need to calculate the probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p].   


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k)\ =\ \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 ways to select *[tex \Large k] things from *[tex \Large n] things where order doesn't matter, i.e. *[tex \Large \frac{n!}{k!(n-k)!}]


For this particular problem *[tex \Large k\ =\ 4], *[tex \Large n\ = 4], and *[tex \Large p\ =\ 0.08]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(4\cr 4\right\)\ =\ \frac{4!}{4!(0)!}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(4)\ =\ \left(1\right\)\(0.08)^4\left(0.92\right)^{0}\ \approx\ 0.000041]


Certainly less precise, but also certainly more illustrative, would be to say, "about a snowball's chance in hell."


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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