Question 361454
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You want the probability of at least 12 successes out of 14 trials where the probability of success on one trial is 0.4.


The probability of exactly *[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)!}]


To get the probability of "at least" some number of successes, then you need to sum the probability of that number of successes plus "that number plus 1" successes plus "that number plus 2" successes...up to the total number of trials :


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{14}\left(\geq12,0.4\right)\ =\ \sum_{i=12}^{14}\left(14\cr\ i\right\)\left(0.4\right)^i\left(0.6\right)^{14\,-\,i}]


Which expands to:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{14}\left(\geq12,0.4\right)\ =\ \left(14\cr12\right\)\left(0.4\right)^{12}\left(0.6\right)^{2}\ +\ \left(14\cr13\right\)\left(0.4\right)^{13}\left(0.6\right)^{1}\ +\ \left(14\cr14\right\)\left(0.4\right)^{14}\left(0.6\right)^{0}]


You have some calculator button punching to do.


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