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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 1, 2, 3, 4, 5, and 6, on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 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...


Solution

Let x be the probability of rolling a 1. The probabilities of rolling a 2, 3, 4, 5, and 6 are 2x, 3x, 4x, 5x, and 6x, respectively.

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

x+2x+3x+4x+5x+6x=1

21x=1

x=\frac{1}{21}

So the probabilities of rolling a 1, 2, 3, 4, 5, and 6 are respectively \frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}, and \frac{6}{21}.

The possible combinations of two rolls that total 7 are: (1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)

The probability of rolling a total of 7 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}+\...

See Also

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