Track / Stage 3 / 10 of 260 #10 of 1964

Problem 10

AMC 10/12, early questions

Combinatorics Difficulty 3.0 Multiple choice

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?
$

Pick one

Official solution

The only way to get a total of 20 $or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of$ \dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28 $ways to choose$ 2 $bills out of$ 8 $. There are$ 12 $ways to choose a twenty and some other non-twenty bill. There is$ 1 $way to choose both twenties, and also$ 1 $way to choose both tens. Adding these up, we find that there are a total of$ 14 $ways to attain a sum of$ 20 $or greater, so there is a total probability of$ \dfrac{14}{28}=\boxed{\textbf{(D) }\frac{1}{2}}$.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.