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2008 AMC 10B Problem 17

2008 AMC 10B Problem 17AMC 10 Step by Step

Name: Date: September 13, 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

Exactly one approval in three independent trials: choose which occasion (3 ways) times 0.7 times 0.3 squared, giving 0.189.

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 $\boxed{\textbf{(B)}\ 0.189}$ .

Why this works

This is the binomial formula $\binom{3}{1} p (1-p)^2$ with $p = 0.7$ . The two ingredients are independence (multiply probabilities along one sequence) and counting the sequences that fit the description (multiply by the number of positions for the approval). Choice (A) is the probability of one particular sequence; the factor of $3$ is what students most often drop.

The trap

Forgetting the factor 3 for which occasion approves (0.063), or computing two approvals and one disapproval (0.441).

Common mistakes

Forgetting the factor 3 for which occasion approves (0.063), or computing two approvals and one disapproval (0.441).
Reading "exactly one" as "at least one," which would give $1 - 0.3^3 = 0.973$ and does not match any choice.

Techniques

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