Question 346244
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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)!}]


But what you need is the probability that there are no more than 2 failed components.  Hence you need the probability of exactly 0 plus the probability of exactly 1 plus the probability of exactly 2.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{11}(\leq 2,0.11)\ =\ \sum_{i=0}^2\left(11\cr\ i\right\)\left(0.11\right)^i\left(0.89\right)^{n\,-\,i}]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{11}(\leq 2,0.11)\ =\ P_{11}(0,0.11)\ +\ P_{11}(1,0.11)\ + P_{11}(2,0.11)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \left(11\cr\, 0\right\)\left(0.11\right)^0\left(0.89\right)^{11}\ +\ \left(11\cr\, 1\right\)\left(0.11\right)^1\left(0.89\right)^{10}\ +\ \left(11\cr\, 2\right\)\left(0.11\right)^2\left(0.89\right)^{9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \left(1\right\)\left(1\right)\left(0.89\right)^{11}\ +\ \left(11\right\)\left(0.11\right)^1\left(0.89\right)^{10}\ +\ \left(55\right\)\left(0.11\right)^2\left(0.89\right)^{9}]


I'll leave you alone to spend some nice quality time with your calculator.


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