2005 AMC 10B Problems/Problem 9: Difference between revisions
Created page with 'An odd sum requires either that the ¯rst die is even and the second is odd or that the ¯rst die is odd and the second is even. The probability is (1/3*1/3)+(2/3*2/3)=1/9+4/9=5/…' |
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An odd sum requires either that the | An odd sum requires either that the first die is even and the second is odd | ||
or that the | or that the first die is odd and the second is even. The probability is | ||
(1/3*1/3)+(2/3*2/3)=1/9+4/9=5/9=D | (1/3*1/3)+(2/3*2/3)=1/9+4/9=5/9=D | ||
Revision as of 22:39, 31 October 2009
An odd sum requires either that the first die is even and the second is odd or that the first die is odd and the second is even. The probability is (1/3*1/3)+(2/3*2/3)=1/9+4/9=5/9=D
