2005 AMC 12B Problem 11

There are $\binom{8}{2} = 28$ equally likely pairs of bills. The sum is $\$20$ or more in these cases: both twenties ($1$ way), one twenty with one of the six smaller bills ($2 \cdot 6 = 12$ ways), or both tens ($1$ way). That is $1 + 12 + 1 = 14$ favorable pairs, so the probability is $\dfrac{14}{28} = \dfrac12.$ Thus, the correct answer is D.