2005 AMC 10B Problem 15

Attempt Problem 15 of the 2005 AMC 10B 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 2005 AMC 10B solutions, or check the answer key.

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15.

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?

14\dfrac14

25\dfrac25

37\dfrac37

12\dfrac12

23\dfrac23

Answer: D
Concepts:basic probabilitycombinationscasework
Difficulty rating: 1370
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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