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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.

 Mar 14, 2019
 #1
avatar+18 
+5

This was a mistake so I just deleted the answer :/

 Mar 14, 2019
edited by DerpyPotato  Mar 14, 2019
edited by DerpyPotato  Mar 15, 2019
 #2
avatar+5655 
+1

\(P[\text{at least 2 1's}] = 1 - P[\text{0 or 1 1's}]\\ P[\text{0 1's}] = \left(\dfrac 5 6\right)^3 = \dfrac{125}{216}\\ P[\text{1 1}] = \dbinom{3}{1}\left(\dfrac 1 6\right)\left(\dfrac 5 6\right)^2 = \dfrac{25}{72}\\ P[\text{at least 2 1's}] = 1 - \dfrac{125}{216}-\dfrac{25}{72} = \dfrac{2}{27}\)

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 Mar 14, 2019

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