Question 584299
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Yep:


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)!}]


For part (a) you want:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(14,0.85)\ =\ \left(15\cr 14\right\)\left(0.85\right)^{14}\left(0.15\right)]


<i><b>Calculation Hint:</b></i> *[tex \LARGE \ \ \left(\ \ n\cr n-1\right\)\ =\ \left(n\cr 1\right\)\ =\ n]


For part (b) you want the probability of exactly 15 plus the result of part (a):



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(14,0.85)\ +\ P_{15}(15,0.85)\ =\ \left(15\cr 14\right\)\left(0.85\right)^{14}\left(0.15\right)\ +\ \left(15\cr 15\right\)\left(0.85\right)^{15}\left(0.15\right)^0]


<i><b>Calculation Hint:</b></i> *[tex \LARGE \ \ \left(n\cr n\right\)\ =\ \left(n\cr 0\right\)\ =\ 1] and *[tex \LARGE a^0\ =\ 1], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(14,0.85)\ +\ P_{15}(15,0.85)\ =\ \left(15\cr 14\right\)\left(0.85\right)^{14}\left(0.15\right)\ +\ \left(0.85\right)^{15}]


Part (c) is just straightforward fill in the numbers, get out the calculator, and crunch noting that the probability of a tail is 1 minus the probability of a head (save for that very, very tiny probability that the coin lands on its edge):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}(3,0.15)\ =\ \left(15\cr \,3\right\)\left(0.15\right)^{3}\left(0.5\right)^{12}]


On the other hand you can save yourself a boatload of arithmetic drugery if you open up a spreadsheet (Excel under Windows or Numbers on a Mac, they both work the same), pick a convenient empty cell, and type in:


Part (a)


=BINOMDIST(14,15,0.85,false)


Part (b)


=BINOMDIST(14,15,0.85,false) + BINOMDIST(15,15,0.85,false)


Alternatively you could enter:


=1 - BINOMDIST(13,15,0.85,true)


Part (c)


=BINOMDIST(3,15,0.15,false)


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