Question 571781
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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)!}]


But you need the sum of the probability of 0 out of 19 plus 1 out of 19 plus 2 out of 19 plus 3 out of 19 plus 4 out of 19, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{19}(\leq4,0.5)\ =\ \sum_{i\,=\,0}^4\left(19\cr\ i\right\)\left(0.5\right)^i\left(0.5\right)^{19\,-\,i}]


If you are a glutton for punishment, you can do all of this arithmetic by hand, or you can open an Excel spreadsheet (or Numbers if you are using a MAC) and type in the following:


=BINOMDIST(4,19,0.5,TRUE)


and when you press return you will get the answer.  You may have to adjust the number of decimal places presented to be able to see the 8 place accuracy required by the problem.  (8 places??? Is your teacher a direct descendant of the Marquis d'Sade?)


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