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2008 AMC 10B Problem 17

2008 AMC 10B Problem 17AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Probability · difficulty 3 of 5 · about 2 min · position P16-20

A poll shows that $70\%$ of all voters approve of the mayor's work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor's work?

A) ${{{0.063}}}$
B) ${{{0.189}}}$
C) ${{{0.233}}}$
D) ${{{0.333}}}$
E) ${{{0.441}}}$

Answer

Key insight

One fixed order, like yes-no-no, has chance 0.7 times 0.3 times 0.3. The single yes can come on any of 3 occasions, so multiply by 3.

Solution

Each occasion is a separate random pick: "approves" with chance $0.7$ , "does not" with chance $0.3$ . List every way to get exactly one "approves".

There are three orders with exactly one approval:

Order	Chance
yes, no, no	$0.7 \times 0.3 \times 0.3 = 0.063$
no, yes, no	$0.3 \times 0.7 \times 0.3 = 0.063$
no, no, yes	$0.3 \times 0.3 \times 0.7 = 0.063$

2. Each order has the same chance, $0.063$ , because it multiplies the same three numbers.
3. These orders cannot happen at the same time, so add them:
$0.063 + 0.063 + 0.063 = 3 \times 0.063 = 0.189.$

The answer is $\boxed{\textbf{(B)}\ 0.189}$ .

Why this works

For separate random picks, the chance of one exact order is the product of the individual chances. "Exactly one" does not say which occasion, so count the orders that fit and multiply. Choice (A) is the chance of just one order; choice (E), $0.441$ , is the chance of exactly two approvals. Listing the orders keeps you from mixing these up.

The trap

Forgetting that the yes can come first, second or third and answering 0.063, which is choice (A).

Concise solution

Each occasion independently yields an approval with probability $0.7$ and a disapproval with probability $0.3$ .

A specific pattern with one approval and two disapprovals, say approve-disapprove-disapprove, has probability
$0.7 \cdot 0.3 \cdot 0.3 = 0.063.$
The single approval could occur on any of the $3$ occasions, and these three patterns are disjoint, so
$3 \cdot 0.063 = 0.189.$

The answer is $\boxed{\textbf{(B)}\ 0.189}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques

Techniques

Set up the equation/formula and compute; no special trick needed