2006 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2006 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

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Concepts:dice (probability)basic probability

Difficulty rating: 1630

21.

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?

$\dfrac{4}{63}$

$\dfrac{1}{8}$

$\dfrac{8}{63}$

$\dfrac{1}{6}$

$\dfrac{2}{7}$

Solution:

Each die shows $k$ with probability $\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}.$

For a total of $7,$ the ordered pairs $(1,6),(2,5),\ldots,(6,1)$ contribute $\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\dfrac{56}{441}=\dfrac{8}{63}.$

Thus, the correct answer is C.

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