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2006 AMC 10B Problems/Problem 21

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Problem

For a particular peculiar pair of dice, the probabilities of rolling , , , , , and , on each die are in the ratio . What is the probability of rolling a total of on the two dice?

\mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7}


Solution

Let be the probability of rolling a . The probabilities of rolling a , , , , and are , , , , and , respectively.

The sum of the probabilities of rolling each number must equal 1, so

So the probabilities of rolling a , , , , , and are respectively \frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}, and .

The possible combinations of two rolls that total are: (1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)

The probability of rolling a total of on the two dice is equal to the sum of the probabilities of rolling each combination.

P = \frac{1}{21}\cdot\frac{6}{21}+\frac{2}{21}\cdot\frac{5}{21}+\frac{3}{21}\cdot\frac{4}{21}+\frac{4}{21}\cdot\frac{3}{21}+\frac{5}{21}\cdot\frac{2}{21}+\frac{6}{21}\cdot\frac{1}{21}=\frac{8}{63} \Rightarrow C

See Also

Preparing for MATHCOUNTS or the AMC contests, and having a tough time with number theory problems? Read Art of Problem Solving's Introduction to Number Theory by Mathew Crawford.
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