2002 AMC 12A Problem 16

Below is the professionally curated solution for Problem 16 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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Concepts:basic probabilitycasework

Difficulty rating: 1630

16.

Tina randomly selects two distinct numbers from the set {1,2,3,4,5},\{1, 2, 3, 4, 5\}, and Sergio randomly selects a number from the set {1,2,,10}.\{1, 2, \ldots, 10\}. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is

25\dfrac{2}{5}

920\dfrac{9}{20}

12\dfrac{1}{2}

1120\dfrac{11}{20}

2425\dfrac{24}{25}

Solution:

Tina's ten pairs have sums 3,4,5,5,6,6,7,7,8,9.3, 4, 5, 5, 6, 6, 7, 7, 8, 9. For a sum s,s, Sergio's number exceeds it with probability 10s10.\dfrac{10 - s}{10}.

The corresponding values of 10s10 - s are 7,6,5,5,4,4,3,3,2,1,7, 6, 5, 5, 4, 4, 3, 3, 2, 1, totaling 40.40. The overall probability is 401010=25.\dfrac{40}{10\cdot 10} = \dfrac{2}{5}.

Thus, the correct answer is A.

Problem 16 in Other Years