Question 1082464
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The probability of <i>k</i> successes in <i>n</i> independent trials where <i>p</i> is the probability of success on any given trial is given by:


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


Where *[tex \Large {{n}\choose{k}}] is the number of combinations of <i>n</i> things taken <i>k</i> at a time.


So you want *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(10,11,0.95)\ =\ {{11}\choose{10}}(0.95)^{10}(1-0.95)^{11-10}]


Hint:  If you are unfamiliar with the *[tex \Large {{n}\choose{k}}] notation, the value can be calculated by *[tex \Large \frac{n!}{k!(n-k)!]


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 \ \ \ \ \ \ \ \ \ \  

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