Difference between revisions of "2005 AMC 10B Problems/Problem 19"

(Redirected page to 2005 AMC 12B Problems/Problem 9)
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== Problem ==
#REDIRECT[[2005 AMC 12B Problems/Problem 9]]
One fair die has faces <math>1</math>, <math>1</math>, <math>2</math>, <math>2</math>, <math>3</math>, <math>3</math> and another has faces <math>4</math>, <math>4</math>, <math>5</math>, <math>5</math>, <math>6</math>, <math>6</math>. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
<math>\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qquad \mathrm{(E)} \frac{2}{3} </math>
== Solution ==
For the sum to be odd, the resulting numbers must be of different parity. The probability that the first die is even and the second die is odd is <math>\frac{1}{3}*\frac{1}{3}=\frac{1}{9}</math> and the probability that the first die is odd and the second die is even is <math>\frac{2}{3}*\frac{2}{3}=\frac{4}{9}</math>. Therefore the probability that the dies have opposing parities (and consequently their sum is odd) is <math>\frac{1}{9}+\frac{4}{9}=\frac{5}{9}\Rightarrow \boxed{\mathrm{(D)}}</math>.
== See Also ==
*[[2005 AMC 10B Problems]]
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 16:28, 12 July 2011

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