Question 363235
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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. you have *[tex \Large n\ =\ 12], *[tex \Large k\ =\ 0], and *[tex \Large p\ =\ 0.28]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(0,0.28)\ =\ \left(12\cr\ 0\right\)\left(0.28\right)^0\left(1\,-\,0.28\right)^{12\,-\,0}]


But since *[tex \LARGE \left(n\cr 0\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{N}] and *[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}], this reduces to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(0,0.28)\ =\ (0.72)^{12}]


Use your calculator and see that it is just under a 2% chance.


b. is a little more complicated.  Fewer than 5 means 0, 1, 2, 3, or 4.  So you need the probability of exactly zero (what you just calcualted above) plus the probability of exactly 1, plus the probability of exactly 2, and so on up to and including 4.  Symbolically:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}\left(<5,0.28\right)\ =\ \sum_{i=0}^5\left(12\cr\ i\right\)\left(0.28\right)^i\left(1\,-\,0.28\right)^{12\,-\,i}]


c. Going at this one straight up, you would calculate the probability of 6, then 7, then 8, and so on up to the probability of all 12, and then add the 7 probability numbers you just calculated.  But having done part b. you can save yourself a bunch of work by realizing that the probability of more than 5 is the same as the probability of at least 5 minus the probability of exactly 5. and further, the probability that fewer than 5 play (the answer to b) plus the probability that at least 5 play is certainty, or 1. So, you need to calculate 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{12}(5,0.28)\ =\ \left(12\cr\ 5\right\)\left(0.28\right)^5\left(0.72\right)^{7}]


Add that to the result of part b. of this problem to get the probability that at most 5 play, and then subtract that sum from 1.


d.  More than half do not play is the same as 7 or more do not play which is the same as fewer than 6 play which is the same as at most 5 play which is the sum of the part b. calculation and the probability of exactly 5 that you worked out in part c.


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