Question 999827
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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)\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE {{n}\choose{k}}] 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 part a, you need the probability of exactly 3, plus the probability of 4, plus...up to 20.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{20}(\geq 3,0.15)\ =\ \sum_{k=3}^{20}\, {{20}\choose{k}}\left(0.15\right)^k\left(0.85\right)^{20\,-\,k}]


Which is the same as 1 minus the sum of the probabilities of 0, 1, and 2 out of 20 trials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{20}(<3,0.15)\ =\ 1\ -\ \sum_{k=0}^{2}\, {{20}\choose{k}}\left(0.15\right)^k\left(0.85\right)^{20\,-\,k}]


and involves one hell of a lot less arithmetic.


for part b:


1 minus the sum of the probabilities of 0 through 29 of 200 trials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{200}(<30,0.15)\ =\ 1\ -\ \sum_{k=0}^{29}\, {{200}\choose{k}}\left(0.15\right)^k\left(0.85\right)^{200\,-\,k}]


Which is way too much ugly arithmetic for me.  Use Excel on a PC or Numbers on a Mac


=1 - BINOMDIST(29,200,0.15,TRUE)


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \