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 combined equals $1/27 + 6/27$ = $\boxed{\textbf{(D)}\ 7/27}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2017 AMC 10B Problems/Problem 9

Problem

Solution

See Also