Question 596859
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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)!}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ 3]


Then, since there are 13 cards in each suit, only three of which are face cards:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p\ =\ \frac{3}{13}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{3}\left(3,\frac{3}{13}\right)\ =\ {{3}\choose{3}}\left(\frac{3}{13}\right)^3\left(\frac{10}{13}\right)^{0}]


=BINOMDIST(3,3,3/13,FALSE) if entered into an Excel (Windows) or Numbers (Mac) spreadsheet will give the answer directly.


But you can also use the following facts to simplify the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {{n}\choose{0}}\ =\ {{n}\choose{n}}\ =\ 1]


for all positive integers and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^0\ =\ 1]


for all real numbers *[tex \LARGE a]


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