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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.

 Feb 28, 2020
 #1
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D=3; C=2; (D nCr C * 5^(D-C) + D nCr (C+1) * 5^(D-C-1)) / 6^D =16 / 216 = 2 / 27

 Feb 28, 2020
 #2
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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

 Feb 28, 2020

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