2005 AMC 10B Problem 15

1. All possible pairs. Choosing $2$ of the $8$ bills can be done in $\frac{8 \cdot 7}{2} = 28$ ways, all equally likely. (Divide by $2$ because picking bill X then bill Y is the same pair as Y then X.)
2. Which pairs reach $\$20$? Without a twenty, the biggest sums are $10 + 10 = 20$ and $10 + 5 = 15$. So a pair works only if it has a twenty, or if it is the two tens.
3. Pairs with a twenty.
- One twenty and one of the $6$ other bills: $2 \cdot 6 = 12$ pairs.
- Both twenties: $1$ pair.
4. The two tens: $1$ pair.
5. Total good pairs: $12 + 1 + 1 = 14$. The probability is $\frac{14}{28} = \frac{1}{2}$.

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

Start by asking which combinations can even reach the target; here that leaves a very short list. Then count each type separately and make sure no pair is in two types. Checking the "just barely" case ($10 + 10 = 20$ exactly) is what catches the trap.

Count the bad pairs instead. A bad pair has no twenty, so both bills come from the $6$ ones, fives and tens: $\frac{6 \cdot 5}{2} = 15$ pairs. But one of those, the two tens, actually reaches $20$, so only $14$ pairs are bad. Good pairs: $28 - 14 = 14$, and $\frac{14}{28} = \frac{1}{2}$.

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