# 2005 AMC 12B Problems/Problem 11

The following problem is from both the 2005 AMC 12B #11 and 2005 AMC 10B #15, so both problems redirect to this page.

## Solution 2

Another way to do this problem is to use complementary counting, i.e. how many ways that the sum is less than 20. Now, you do not have to consider the 2 twenties, so you have 6 bills left. $\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15$ ways. However, you counted the case when you have 2 tens, so you need to subtract 1, and you get 14. Finding the ways to get 20 or higher, you subtract 14 from 28 and get 14. So the answer is $\dfrac{14}{28} = \boxed{\mathrm{(D)}\ \dfrac{1}{2}}$.

## Solution 3

If you think about the problem, it's the same as trying as trying to pick from one dollar bill, one five dollar bill, one ten dollar bill, and one twenty dollar bill and trying to get a single bill that is more than or equal to ten dollars. We simply have the half all the values because the expected value gets halved along with the probability of choosing the bills, hence being the same problem. In doing so, two of the four options listed above are greater than or equal to ten dollars, so the answer is $\boxed{\mathrm{(D)}\ \dfrac{1}{2}}$.