2005 AMC 10B Problem 15

There are $\binom{8}{2} = 28$ equally likely pairs of bills.

The largest sum without a twenty and without two tens is $10 + 5 = 15$, so a sum of $20$ or more requires either at least one twenty, or both tens.

• At least one twenty: pairs drawn entirely from the six non-twenty bills number $\binom{6}{2} = 15$, so $28 - 15 = 13$ pairs include a twenty.
• No twenty but both tens: exactly $1$ pair.

These cases do not overlap, so there are $13 + 1 = 14$ favorable pairs, and the probability is $\frac{14}{28} = \frac{1}{2}$.

The answer is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.

Identify the threshold first: which combinations can even reach $20$? That turns a "sum" question into a short list of qualifying types. "At least one twenty" is counted fastest by complement, and the leftover case (two tens) is a single pair; keeping the cases disjoint avoids double counting.

Direct count: a twenty with a non-twenty ($2 \cdot 6 = 12$ pairs), both twenties ($1$), both tens ($1$): $14$ pairs, same result.

• Forgetting the pair of two tens (giving 13/28) or double counting the pair of two twenties.
• Treating "twenty with another bill" as $2 \cdot 7 = 14$ pairs, which counts the two-twenties pair twice.