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

(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 ¯rst die is even and the second is odd
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An odd sum requires either that the first die is even and the second is odd
or that the ¯rst 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
 
(1/3*1/3)+(2/3*2/3)=1/9+4/9=5/9=D

Revision as of 23:39, 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