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

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== Problem == | == Problem == | ||

+ | 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? | ||

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+ | <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> | ||

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== Solution == | == 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>. | ||

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== See Also == | == See Also == | ||

*[[2005 AMC 10B Problems]] | *[[2005 AMC 10B Problems]] | ||

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+ | [[Category:Introductory Combinatorics Problems]] |

## Revision as of 14:15, 7 February 2009

## Problem

One fair die has faces , , , , , and another has faces , , , , , . The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

## 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 and the probability that the first die is odd and the second die is even is . Therefore the probability that the dies have opposing parities (and consequently their sum is odd) is .