2005 AMC 10B Problem 15

P11-15~3 minPrint

Basic Probability Basic Counting Split into exhaustive cases and handle each Count the complement and subtract from the total

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?

Show solution without answering

Key insight

Only two kinds of pairs reach 20 dollars: any pair with a twenty, or the two tens. That is 14 of the 28 possible pairs.

Solution

1. All possible pairs. Choosing $2$ of the $8$ bills can be done in $\frac{8 \cdot 7}{2} = 28$ ways, all equally likely. (Divide by $2$ because picking bill X then bill Y is the same pair as Y then X.)
2. Which pairs reach $\$20$ ? Without a twenty, the biggest sums are $10 + 10 = 20$ and $10 + 5 = 15$ . So a pair works only if it has a twenty, or if it is the two tens.
3. Pairs with a twenty.
- One twenty and one of the $6$ other bills: $2 \cdot 6 = 12$ pairs.
- Both twenties: $1$ pair.
4. The two tens: $1$ pair.
5. Total good pairs: $12 + 1 + 1 = 14$ . The probability is $\frac{14}{28} = \frac{1}{2}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ .

Why this works

Start by asking which combinations can even reach the target; here that leaves a very short list. Then count each type separately and make sure no pair is in two types. Checking the "just barely" case ( $10 + 10 = 20$ exactly) is what catches the trap.

Alternative approach

Count the bad pairs instead. A bad pair has no twenty, so both bills come from the $6$ ones, fives and tens: $\frac{6 \cdot 5}{2} = 15$ pairs. But one of those, the two tens, actually reaches $20$ , so only $14$ pairs are bad. Good pairs: $28 - 14 = 14$ , and $\frac{14}{28} = \frac{1}{2}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

There are $\binom{8}{2} = 28$ equally likely pairs of bills.

The largest sum without a twenty and without two tens is $10 + 5 = 15$ , so a sum of $20$ or more requires either at least one twenty, or both tens.

At least one twenty: pairs drawn entirely from the six non-twenty bills number $\binom{6}{2} = 15$ , so $28 - 15 = 13$ pairs include a twenty.
No twenty but both tens: exactly $1$ pair.

These cases do not overlap, so there are $13 + 1 = 14$ favorable pairs, and the probability is $\frac{14}{28} = \frac{1}{2}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques