Question 333206
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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)!}]


For part a, your number of trials, *[tex \Large n] is 12, your desired number of successes, *[tex \Large k] is 7, and your probability of success on an individual trial, *[tex \Large p] is *[tex \Large \frac{2}{6}\ =\ \frac{1}{3}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}\left(7,\frac{1}{3}\right)\ =\ \left(12\cr\ 7\right\)\left(\frac{1}{3}\right)^7\left(\frac{2}{3}\right)^5]


The other problem is worked the same way, except your number of trials, *[tex \Large n] is 12, your desired number of successes, *[tex \Large k] is 4, and your probability of success on an individual trial, *[tex \Large p] is *[tex \Large \frac{4}{6}\ =\ \frac{2}{3}], and you want to compute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}\left(4,\frac{2}{3}\right)]


I'll leave you alone to spend some 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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