Question 276332
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There are 5 ways out of 6 possibilities that any one of the dice will NOT be a 4.  Therefore the probability of NO 4s being rolled is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{5}{6}\right)^5]


and therefore the probability of at least one 4 is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \left(\frac{5}{6}\right)^5]


You can punch your own calculator buttons, but it is roughly 60 percent in favor of at least one 4.


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