Question 761278
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No.


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)\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE {{n}\choose{k}}] 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 need *[tex \LARGE P_{15}(0,0.4)\ =\ {{15}\choose{0}}\left(0.4\right)^0\left(0.6\right)^{15}]


I'll let you do your own arithmetic.  (hint: n choose 0 = 1)


For part b, the straight forward approach is to calculate the above for *[tex \LARGE k\ =\ 3], then *[tex \LARGE k\ =\ 4], then 5, 6, ... and so on up to 15. And then you have to add up all of your answers for these results.  Lots of VERY nasty arithmetic.  Looks like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(\geq k,p)\ =\ \sum_{i\,=\,k}^n\,{{n}\choose{i}}\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i}]



Better approach:  The probability of AT LEAST 3 is equal to 1 minus the probability of LESS THAN 3.  The probability of less than 3 is the probability of zero (which is what you just calculated in part a) plus the probability of 1 plus the probability of 2.  Don't forget to subtract the sum from 1.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(<k,p)\ =\ 1\ -\ \sum_{i\,=\,0}^{k\,-\,1}\,{{n}\choose{i}}\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i}]


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