Question 232598
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Presuming two fair, 6-sided dice, each numbered 1 through 6:


You can achieve one of the following results on any given roll: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.  But there is only 1 way to get a 2, 2 ways to get a 3, 3 ways to get a 4, up to 6 ways to get a 7, then 5 ways to get 8, and so on.  Total is 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36 possible outcomes.


The prime numbers from 2 through 12 inclusive are: 2, 3, 5, 7, and 11. There is 1 way to get a 2, 2 ways for a 3, 4 ways for a 5, 6 ways for a 7 and 2 ways for 11.  Total of 1 + 2 + 4 + 6 + 2 = 15 successful outcomes out of 36 possible outcomes.


You can take it from here.



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