2005
Problem - 1774
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?
There are totally $C_8^2=28$ ways to pick two bills out of $8$.
In order to make their sum $20$ or more, either one of them is the twenty bill or both bills are tens. There are $C_2^1\times C_6^1 = 12$ ways to pick one twenty and one non-twenty, $1$ way to pick two twenties, and $1$ way to pick two tens. Therefore the answer is $$\frac{12+1+1}{28}=\boxed{\frac{1}{2}}$$