Question 760642
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What you need is the probability of meeting 1 plus the probability of meeting 2 plus...all the way up through 8.


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 you need to calculate for *[tex \LARGE k\ =\ 1], *[tex \LARGE n\ =\ 8], and *[tex \LARGE p\ =\ 0.11], then *[tex \LARGE k\ =\ 2], *[tex \LARGE n\ =\ 8], and *[tex \LARGE p\ =\ 0.11], then *[tex \LARGE k\ =\ 3], and so on up through 8, then add all of the results.


There is a much easier way though.  You see, the probability of "at least 1" is the same as 1 minus the probability of none.  So calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n;k,p)\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE k\ =\ 0], *[tex \LARGE n\ =\ 8], and *[tex \LARGE p\ =\ 0.89] and then subtract the result from 1. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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