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2019 AMC 10B Problem 17

2019 AMC 10B Problem 17AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Probability · difficulty 3 of 5 · about 4 min · position P16-20

A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k=1,2,3,\ldots.$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

A) $\frac{1}{4}$
B) $\frac{2}{7}$
C) $\frac{1}{3}$
D) $\frac{3}{8}$
E) $\frac{3}{7}$

Answer

Key insight

Red higher and green higher are equally likely. The only other result is a tie, which has probability 1/4 + 1/16 + 1/64 + ... = 1/3.

Solution

The two balls are tossed the same way, so neither has an advantage. Use that instead of adding up a big table.

1. There are three possible results: red in a higher bin, green in a higher bin, or both in the same bin (a tie).
2. By symmetry, "red higher" and "green higher" have the same probability. Call it $p$ . Then
$p + p + (\text{probability of a tie}) = 1.$
3. Probability of a tie. Both balls land in bin $1$ with probability $\frac12 \cdot \frac12 = \frac14$ . Both in bin $2$ : $\frac14 \cdot \frac14 = \frac1{16}$ . Both in bin $3$ : $\frac18 \cdot \frac18 = \frac1{64}$ . And so on. So
$\text{tie} = \frac14 + \frac1{16} + \frac1{64} + \cdots$
4. Add up that sum. Call it $S$ . Multiplying by $4$ gives $4S = 1 + \frac14 + \frac1{16} + \cdots = 1 + S$ . So $3S = 1$ and $S = \frac13$ .
5. Back to step 2: $2p + \frac13 = 1$ , so $2p = \frac23$ and $p = \frac13$ .

The answer is $\boxed{\textbf{(C)}\ \frac{1}{3}}$ .

Why this works

When two things are chosen the same random way and compared, "first is bigger" and "second is bigger" are equally likely. So you only have to find the chance of a tie, and a tie is easy because both balls go to the same bin. Remember the pattern: answer $= \frac12(1 - \text{tie})$ .

Alternative approach

Count directly. If green lands in bin $k$ (probability $\frac{1}{2^k}$ ), red is higher when it lands in bin $k+1$ , $k+2$ , and so on, with probability $\frac{1}{2^{k+1}} + \frac{1}{2^{k+2}} + \cdots = \frac{1}{2^k}$ . So
$p = \frac12 \cdot \frac12 + \frac14 \cdot \frac14 + \frac18 \cdot \frac18 + \cdots = \frac14 + \frac1{16} + \frac1{64} + \cdots = \frac13.$

The trap

Answering 1/2 by forgetting that both balls can land in the same bin.

Concise solution

The two balls follow the same distribution independently, so "red higher" and "green higher" have the same probability $p$ . The remaining possibility is a tie:
$p + p + P(\text{tie}) = 1.$

Both balls land in bin $k$ with probability $2^{-k} \cdot 2^{-k} = 4^{-k}$ , so
$P(\text{tie}) = \sum_{k=1}^{\infty} 4^{-k} = \frac{1/4}{1 - 1/4} = \frac13.$

Therefore $2p = 1 - \frac13 = \frac23$ and $p = \frac13$ .

The answer is $\boxed{\textbf{(C)}\ \frac{1}{3}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.3)

Techniques

Count the complement and subtract from the total · Exploit symmetry to reduce work or pair up objects