Question 667012
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where *[tex \Large p] is the probability of success on any given trial is given by:


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


Where *[tex \LARGE {{n}\choose{k}}] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


You want the probability of getting either 4 or 5, so calculate the probability of getting all 5 right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5(5,0.5)\ =\ {{5}\choose{5}}\left(0.5\right)^5\left(1\,-\ 0.5\right)^{5\,-\,5}]


then calculate the probability of getting exactly 4 right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5(4,0.5)\ =\ {{5}\choose{4}}\left(0.5\right)^4\left(1\,-\ 0.5\right)^{5\,-\,4}]


And then add the two values.  Note: *[tex \LARGE {{n}\choose{n}}\ =\ 1], *[tex \LARGE {{n}\choose{n-1}}\ =\ n], anything to the zero power is 1, and anything to the 1 power is itself.


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