2017 AMC 10B Problems/Problem 9

Problem

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?

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

Solution

There are two ways that the contestant can win.

Case 1: They guess all three right. This can only happen $1/3 * 1/3 * 1/3 = 1/27$ of the time.

Case 2: They guess only two right. We pick one of the questions to get wrong, $3$ , and this can happen $1/3 * 1/3 * 2/3$ of the time. Thus, $2/27 * 3$ = $6/27$ .

So in total the two cases equal $1/27 + 6/27$ = $\boxed{\textbf{(D)}\ 7/27}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2017 AMC 10B Problems/Problem 9

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