Question 345596
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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)!}]


For your situation, *[tex \Large p\ =\ 0.8], but you didn't specify how many flights were observed nor exactly how many should be on time, so I can't help you with *[tex \Large n] and *[tex \Large k]


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