Question 320257
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Part a.  You want the probability of *[tex \Large k] successes out of *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k)\ =\ \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 computed by *[tex \LARGE \frac{n!}{k!(n\ -\ k)!}].


For this particular problem:  


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  k\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  n\ =\ 15]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p\ =\ \frac{1}{4}]


Hence you need to calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(5)\ =\ \left(15\cr \,5\right\)\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^{10}]


where *[tex \LARGE \left(15\cr \,5\right\)] is the number of combinations of *[tex \LARGE 15] things taken *[tex \LARGE 5] at a time and is computed by *[tex \LARGE \frac{15!}{5!(10)!}].


I'll leave the calculator work to you.


Part b.  The probability of getting less than 6 correct is the sum of the probability of getting exactly 0 plus exactly 1 plus...plus exactly 5.  In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(<6)\ =\ \sum_{k=0}^5\left(15\cr \,k\right\)\left(\frac{1}{4}\right)^k\left(\frac{3}{4}\right)^{15\ -\ k}]


Again, I'll leave the calculator work to you.


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