2009 AMC 10A Problems/Problem 22

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Problem

Two cubical dice each have removable numbers $1$ through $6$. The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is $7$?

$\mathrm{(A)}\ \frac{1}{9} \qquad \mathrm{(B)}\ \frac{1}{8} \qquad \mathrm{(C)}\ \frac{1}{6} \qquad \mathrm{(D)}\ \frac{2}{11} \qquad \mathrm{(E)}\ \frac{1}{5}$

Solution

At the moment when the numbers are in the bag, imagine that each of them has a different color. Clearly the situation is symmetric at this moment. Hence after we draw them, attach them and throw the dice, the probability of getting some pair of colors is the same for any two colors.

There are ${12 \choose 2} = 66$ ways how to pick two of the colors. We now have to count the ways where the two chosen numbers will have sum $7$.

Sum $7$ can be obtained as $1+6$, $2+5$, or $3+4$. Each number in the bag has two different colors, hence each of these three options corresponds to four pairs of colors.

Out of the $66$ pairs of colors we can get when throwing the dice, $3\cdot 4=12$ will give us the sum $7$. Hence the probability that this will happen is $\frac{12}{66} = \boxed{\frac 2{11}}$.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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