All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2

There are total of 2+2+2+2=8 bills. The sum of 2 bills will be 20 or more if we pick two $10 bills or $20 bill plus any bill for the other one (which means $20 bill and not $20 bill and two $20 dollar bills).

\(\frac{2}{8}*\frac{1}{7}+2*\frac{2}{8}*\frac{6}{7}+\frac{2}{8}*\frac{1}{7}=\frac{1}{2}\) (we multiply by 2 the second term as it represents the probability of $20/Not $20 scenario which can occur in 2 ways $20/Not $20 and Not $20/$20).

Answer: D.

Or another way:

Find the probability of opposite event and subtract it from 1.

Opposite event would be the sum being less than $20 which means the probability of picking any 2 bills but twenties (so any from 6) minus the case when we pick two $10 bills (1 case): \(\frac{C^2_6-1}{C^2_8}=\frac{1}{2}\).

Answer: D.