Question 1131108
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The probability of exactly *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p] is given by:


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


You want the probability of 2 successes in 3 trials with a probability of success per trial of *[tex \Large \frac{1}{6}].  Plug in your numbers and do the arithmetic.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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