Question 990557
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There is 1 way to make a 2, 3 ways to make four, 5 ways to make six, 5 ways to make 8, 3 ways to make ten, and 1 way to make 12, for a total of 18 ways to make an even number.


There are 2 ways to make three, 4 ways to make five, 6 ways to make seven, 4 ways to make nine, and 2 ways to make eleven, for a total of 18 ways to make an odd number.


Hence, the probability of an even total on any given roll is  *[tex \Large 0.5]


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)!}]


You need the probability of 1 success in 7 trials where the probability of success on any given trial is 0.5, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_7(1,0.5)\ =\ {{7}\choose{1}}\left(0.5\right)^1\left(1\,-\,0.5\right)^{7\,-\,1}]


Get out your calculator.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \