Question 1171688
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The probability of *[tex \Large k] successes in *[tex \Large n] independent trials where the probability of success on any trial is *[tex \Large p] is given by:


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


For 0 or 6 successes out of 10 with a probability of 30%


You need to calculate: *[tex \Large P(0,10,0.3)\ +\ P(6,10,0.3)]. You can do your own arithmetic.

																
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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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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