Question 324753
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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)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE \left(n\cr k\right\)] 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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(9,0.97)\ =\ \left(10\cr\  9\right\)\left(0.97\right)^9\left(0.03\right)^{1}]


Hint: *[tex \Large  \left(\ \,n\cr n-1\right\)\ =\ \left(n\cr 1\right\)\ =\ n]


The rest of the arithmetic is yours to do.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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