2002 AMC 12A Problem 16

Attempt Problem 16 of the 2002 AMC 12A 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 2002 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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}

Answer: A
Concepts:basic probabilitycasework
Difficulty rating: 1630
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 15#15
Full Exam

Problem 16 in Other Years