2006 AMC 10B Problem 21

Attempt Problem 21 of the 2006 AMC 10B below, then check your answer against the professionally curated solution 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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21.

For a particular peculiar pair of dice, the probabilities of rolling 1,2,3,4,5,1,2,3,4,5, and 66 on each die are in the ratio 1:2:3:4:5:6.1:2:3:4:5:6. What is the probability of rolling a total of 77 on the two dice?

463\dfrac{4}{63}

18\dfrac{1}{8}

863\dfrac{8}{63}

16\dfrac{1}{6}

27\dfrac{2}{7}

Answer: C
Concepts:dice (probability)basic probability
Difficulty rating: 1630
Solution:

Each die shows kk with probability k1+2++6=k21.\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}.

For a total of 7,7, the ordered pairs (1,6),(2,5),,(6,1)(1,6),(2,5),\ldots,(6,1) contribute 16+25+34+43+52+61212=56441=863. \begin{aligned} &\scriptsize\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}\\ &=\dfrac{56}{441}=\dfrac{8}{63}. \end{aligned}

Thus, the correct answer is C.

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