Question 474197
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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\)\left(p\right)^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)!}], and which can be calculated in MS Excel using:


=COMBIN(n,k)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p\ =\ 0.33]


Part a)


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


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(3,0.33)\ =\ \left(12\cr\ 3\right\)\left(0.33\right)^3\left(0.67\right)^{9}]


Which can be calculated using Excel:


=BINOMDIST(3, 12, 0.33, FALSE)


Part b)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ \geq\ 4]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(\geq{4},0.33)\ =\ \sum_{i\,=\,4}^{12}\ \left(12\cr\ i\right\)\left(0.33\right)^i\left(0.67\right)^{12\,-\,i}]


But this is equal to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P_{12}(\leq{3},0.33)\ =\ 1\ -\ \sum_{i\,=\,0}^3\ \left(12\cr\ i\right\)\left(0.33\right)^i\left(0.67\right)^{12\,-\,i}]


Which can be calculated using Excel:


=1 - BINOMDIST(3, 12, 0.33, TRUE)


Part c)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ \leq\ 2]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ P_{12}(\leq{2},0.33)\ =\ \sum_{i\,=\,0}^2\ \left(12\cr\ i\right\)\left(0.33\right)^i\left(0.67\right)^{12\,-\,i}]


Which can be calculated using Excel:


=BINOMDIST(2, 12, 0.33, TRUE)


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