Question 1180991
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The probability of *[tex \Large x] successes in *[tex \Large n] independent trials where the probability of success on any given trial is *[tex \Large p] is given by *[tex \Large P(x,n,p)\ =\ {{n}\choose{x}}\,\(p\)^x\,\(1-p\)^{n-x}]


You want the probability of 3 successes in 18 trials where the probability of success on any given trial is 0.25. You can do your own arithmetic.


The mean of a binomial distribution is *[tex \Large \mu\ =\ np]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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{{n}\choose{r}}
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