If Michael rolls three fair dice, what is the probability that he will roll at least two 1's? Express your answer as a common fraction.
D=3; C=2; (D nCr C * 5^(D-C) + D nCr (C+1) * 5^(D-C-1)) / 6^D =16 / 216 = 2 / 27
Here's another way of looking at the problem:
Probability he will roll no 1's = (5/6)3.
Probability he will roll exactly one 1 = 3*(1/6)*(5/6)2.
Hence probability he will roll at least two 1's = 1 - ((5/6)3 + 3*(1/6)*(5/6)2) = 2/27
