2005 AMC 10B Problem 15

Question

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

There are

\binom82

equally likely pairs of bills

A sum of

\$20

or more needs both tens, both twenties, or a twenty with any other bill

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.

Problem 15 in other years