Difference between revisions of "2005 AMC 10B Problems/Problem 9"
| Line 1: | Line 1: | ||
An odd sum requires either that the first die is even and the second is odd | 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 | 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 | + | <math>(1/3*1/3)+(2/3*2/3)</math>=<math>1/9+4/9=5/9</math>='''D''' |
Revision as of 22:48, 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
=
=D
