2005 AMC 10B Problem 15

Below is the professionally curated solution for Problem 15 of the 2005 AMC 10B, 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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Concepts:basic probability combinations casework

Difficulty rating: 1370

15.

An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $\$20$ or more?

$\dfrac14$

$\dfrac25$

$\dfrac37$

$\dfrac12$

$\dfrac23$

Solution:

There are $\binom82 = 28$ equally likely pairs.

A sum of at least $\$20$ comes from both twenties ( $1$ way), a twenty paired with any of the six smaller bills ( $2 \cdot 6 = 12$ ways), or both tens ( $1$ way).

The probability is $\dfrac{1 + 12 + 1}{28} = \dfrac{14}{28} = \dfrac12.$

Thus, D is the correct answer.

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