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2006 AMC 12B Problems/Problem 17

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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 18 \qquad\mathrm{(C)}\ \frac 8{63}\qquad\mathrm{(D)}\ \frac 16\qquad\mathr...

Solution

The probability of getting an x on one of these dice is \frac{x}{21}.

The probability of getting 1 on the first and 6 on the second die is \frac 1{21}\cdot\frac 6{21}. Similarly we can express the probabilities for the other five ways how we can get a total 7. (Note that we only need the first three, the other three are symmetric.)

Summing these, the probability of getting a total 7 is: 2\cdot\left(\frac 1{21}\cdot\frac 6{21}+\frac 2{21}\cdot\frac 5{21}+\frac 3{21}\cdot\frac 4{21}\right)= \frac{56}{441}=\boxed...

See also

2006 AMC 12B (ProblemsResources)
Preceded by
Problem 16
Followed by
Problem 18
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