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2008 AMC 10B Problem 20

Below is the professionally curated solution for Problem 20 of the 2008 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

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Concepts:dice (probability)systematic listing

Difficulty rating: 1510

20.

The faces of a cubical die are marked with the numbers 1,2,2,3,3,1,2,2,3,3, and 4.4. The faces of a second cubical die are marked with the numbers 1,3,4,5,6,1,3,4,5,6, and 8.8. Both dice are thrown. What is the probability that the sum of the two top numbers will be 5,5, 7,7, or 9?9?

518\dfrac{5}{18}

718\dfrac{7}{18}

1118\dfrac{11}{18}

34\dfrac{3}{4}

89\dfrac{8}{9}

Solution:

Of the 3636 equally likely outcomes, the pairs giving sum 55 are (1,4),(2,3),(2,3),(4,1),(1,4),(2,3),(2,3),(4,1), which is 44 outcomes.

Sum 77 comes from (1,6),(2,5),(2,5),(3,4),(3,4),(4,3),(1,6),(2,5),(2,5),(3,4),(3,4),(4,3), which is 6,6, and sum 99 from (1,8),(3,6),(3,6),(4,5),(1,8),(3,6),(3,6),(4,5), which is 4.4.

The probability is 4+6+436=1436=718.\tfrac{4+6+4}{36}=\tfrac{14}{36}=\tfrac{7}{18}.

Thus, the correct answer is B.

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