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2005 AMC 10B Problem 15

Problem 15 of 25IntermediateCounting & Probability

An envelope contains eight bills: 22 ones, 22 fives, 22 tens, and 22 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $20\$20 or more?

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Solution

There are (82)=28\binom82 = 28 equally likely pairs. A sum of at least $20\$20 comes from both twenties (11 way), a twenty paired with any of the six smaller bills (26=122 \cdot 6 = 12 ways), or both tens (11 way). The probability is 1+12+128=1428=12. \dfrac{1 + 12 + 1}{28} = \dfrac{14}{28} = \dfrac12. Thus, D is the correct answer.

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Concepts: basic probability · combinations · casework

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.