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2017 AMC 10B Problem 9

2017 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Probability · difficulty 2 of 5 · about 3 min · position P1-10

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

A) $\frac{1}{27}$
B) $\frac{1}{9}$
C) $\frac{2}{9}$
D) $\frac{7}{27}$
E) $\frac{1}{2}$

Answer

Key insight

There are 27 equally likely ways to answer. Exactly two right happens in 3 times 2 = 6 ways, all three right in 1 way: 7 out of 27.

Solution

Each question has $3$ choices, so there are $3 \cdot 3 \cdot 3 = 27$ ways to answer the quiz, all equally likely. Count the ways that win.

On each question, $1$ choice is right and $2$ are wrong.
Exactly two right. First pick which question is the wrong one: $3$ ways (the first, second or third). That question has $2$ wrong choices, and the other two questions must be answered correctly. So $3 \cdot 2 = 6$ ways.
All three right. Only $1$ way.
Winning ways: $6 + 1 = 7$ out of $27$ .

The answer is $\boxed{\textbf{(D)}\ \frac{7}{27}}$ .

Why this works

"Two or more" splits into the separate cases "exactly two" and "exactly three." In each case, first choose where the wrong answers go, then how many ways they can be wrong. Counting ways out of $27$ keeps the work in whole numbers instead of fractions.

Alternative approach

Count the losing ways and subtract. Losing means $0$ or $1$ right.

Zero right: every question has $2$ wrong choices, so $2 \cdot 2 \cdot 2 = 8$ ways.
Exactly one right: pick the right question ( $3$ ways), and the other two are each wrong in $2$ ways, so $3 \cdot 2 \cdot 2 = 12$ ways.

Losing ways: $8 + 12 = 20$ , so winning ways are $27 - 20 = 7$ , giving $\frac{7}{27}$ .

The trap

Forgetting that the one wrong answer can be on any of the 3 questions, which counts only 2 + 1 = 3 winning ways and gives 3/27.

Concise solution

Each question is answered correctly with probability $\frac13$ and incorrectly with probability $\frac23$ , independently. There are $3^3 = 27$ equally likely answer patterns.

Exactly two right: choose which question is missed ( $3$ ways), then that question has $2$ wrong choices while the other two are forced. That is $3 \cdot 2 = 6$ patterns.

All three right: $1$ pattern.

Winning patterns: $6 + 1 = 7$ out of $27$ .

The answer is $\boxed{\textbf{(D)}\ \frac{7}{27}}$ .

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 (§2.2)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed