Question 1118749
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any one trial is the binomial distribution,


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


Your *[tex \Large n] is 6 and your *[tex \Large p] is 0.6.


For the first part, *[tex \Large k\ =\ 3]


For the second part, you need the sum of the probabilities when *[tex \Large k\ =\ 3], *[tex \Large k\ =\ 4], *[tex \Large k\ =\ 5], and *[tex \Large k\ =\ 6], which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(6,\geq 3,0.6)\ =\ \sum_{k=3}^6\,{{6}\choose{k}}\,(0.6)^k(0.4)^{6-k}]


Get out your calculator, you have a bunch of arithmetic to do.  
								
								
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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