Question 656790
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You want the binomial distribution:


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)!}]


So for the probability of exactly 2,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(2,0.05)\ =\ {{12}\choose{2}}\left(0.05\right)^2\left(1\,-\,0.05\right)^{12\,-\,2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(66\right)\left(0.05\right)^2\left(0.95)^{10}]


A little calculator work is all that is left.


At least two is a little more complicated.  Here you need the sum of exactly 2 plus exactly 3 plus exactly 4...and so on.  A long and tedious computation indeed.  Fortunately there is a simpler method.  The probablity of 0 plus P(1) plus P(2), etc., all the way up to 12 is equal to 1.  So if you want the sum of P(2) plus P(3), etc. all the way to 12, calculate P(0) plus P(1) and subtract from 1.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(0,0.05)\ =\ {{12}\choose{0}}\left(0.05\right)^0\left(1\,-\,0.05\right)^{12\,-\,0}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\right)\left(1\right)\left(0.95)^{12}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(1,0.05)\ =\ {{12}\choose{1}}\left(0.05\right)^1\left(1\,-\,0.05\right)^{12\,-\,1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(12\right)\left(0.05\right)\left(0.95)^{11}]


So you need to punch:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (0.95)^{12}\ +\ (12)(0.05)(0.95)^{11}]


Into your calculator and then subtract the result from 1.  Still ugly, but not so bad a mother couldn't love it.


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