Difference between revisions of "2005 AMC 10B Problems/Problem 9"
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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
