Question 975208
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Exactly 5 threes:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}\left(5,\frac{1}{6}\right)\ =\ {{15}\choose{5}}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^{10}]



At most 5 threes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}\left(\leq 5,\frac{1}{6}\right)\ =\ \sum_{k=0}^{5}\,{{15}\choose{k}}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{15-k}]



At least 5 threes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{15}\left(\geq 5,\frac{1}{6}\right)\ =\ 1\ -\ \sum_{k=0}^{4}\,{{15}\choose{k}}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{15-k}]


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

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