There are \[
\binom{8}{2} = \frac{8!}{6!\cdot 2!} = 28
\]ways to choose the bills. A sum of at least $\$20$ is obtained by choosing  both $\$20$ bills, one of the  $\$20$ bills and one of the six smaller bills, or both $\$10$ bills.   Hence the probability is \[
\frac{ 1 + 2\cdot 6 + 1}{28}=\frac{14}{28}=\boxed{\frac{1}{2}}.
\]
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?
{'answer': '\\frac{1}{2}', 'subject': 'Counting & Probability', 'level': 5, 'unique_id': 'test/counting_and_probability/483.json'}