2005 AMC 10B Problems/Problem 19

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Problem

One fair die has faces $1$ , $1$ , $2$ , $2$ , $3$ , $3$ and another has faces $4$ , $4$ , $5$ , $5$ , $6$ , $6$ . The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qqu...$

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 $\frac{1}{3}*\frac{1}{3}=\frac{1}{9}$ and the probability that the first die is odd and the second die is even is $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$ . Therefore the probability that the dies have opposing parities (and consequently their sum is odd) is $\frac{1}{9}+\frac{4}{9}=\frac{5}{9}\Rightarrow \boxed{\mathrm{(D)}}$ .

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2005 AMC 10B Problems/Problem 19

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