2002 AMC 10A Problem 24

Below is the professionally curated solution for Problem 24 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:basic probabilitycasework

Difficulty rating: 1900

24.

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 1010 equally likely pairs give 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}.

Averaging the winning probability over the ten pairs, the total is 7+6+5+5+4+4+3+3+2+1100=40100=25.\dfrac{7+6+5+5+4+4+3+3+2+1}{100}=\dfrac{40}{100}=\dfrac{2}{5}.

Thus, the correct answer is A.

Problem 24 in Other Years