2005 AMC 10B Problem 9

Below is the professionally curated solution for Problem 9 of the 2005 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10B solutions, or check the answer key.

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Concepts:dice (probability)parityindependent events

Difficulty rating: 1240

9.

One fair die has faces 1,1,2,2,3,31, 1, 2, 2, 3, 3 and another has faces 4,4,5,5,6,6.4, 4, 5, 5, 6, 6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

13\dfrac13

49\dfrac49

12\dfrac12

59\dfrac59

23\dfrac23

Solution:

The first die is odd (a 11 or 33) with probability 23\dfrac23 and even with probability 13.\dfrac13. The second die is odd (a 55) with probability 13\dfrac13 and even with probability 23.\dfrac23.

The sum is odd when the two parities differ: 1313+2323=19+49=59. \dfrac13 \cdot \dfrac13 + \dfrac23 \cdot \dfrac23 = \dfrac19 + \dfrac49 = \dfrac59.

Thus, D is the correct answer.

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