Difference between revisions of "2017 AMC 10B Problems/Problem 9"

Revision as of 03:14, 4 December 2020

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 1

There are two ways the contestant can win.

Case 1: The contestant guesses all three right. This can only happen $\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}$ of the time.

Case 2: The contestant guesses only two right. We pick one of the questions to get wrong, $3$ , and this can happen $\frac{1}{3} * \frac{1}{3} * \frac{2}{3}$ of the time. Thus, $\frac{2}{27} * 3$ = $\frac{6}{27}$ .

So, in total the two cases combined equals $\frac{1}{27} + \frac{6}{27}$ = $\boxed{\textbf{(D)}\ \frac{7}{27}}$ .

Solution 2 (complementary counting)

Complementary counting is good for solving the problem and checking work if you solved it using the method above.

There are two ways the contestant can lose.

Case 1: The contestant guesses zero questions correctly.

The probability of guessing incorrectly for each question is $\frac{2}{3}$ . Thus, the probability of guessing all questions incorrectly is $\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}$ .

Case 2: The contestant guesses one question correctly. There are 3 ways the contestant can guess one question correctly since there are 3 questions. The probability of guessing correctly is $\frac{1}{3}$ so the probability of guessing one correctly and two incorrectly is $3* \frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{9}$ .

The sum of the two cases is $\frac{8}{27} + \frac{4}{9} = \frac{20}{27}$ . This is the complement of what we want to the answer is $1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}$

Video Solution

https://youtu.be/XYeexmAyVzQ

~savannahsolver

Video Solution 2

https://youtu.be/XRfOULUmWbY?t=482

~IceMatrix

@@ Line 31: / Line 31: @@
 ~savannahsolver
+==Video Solution 2==
+https://youtu.be/XRfOULUmWbY?t=482
+~IceMatrix
 ==See Also==
 {{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}

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

Page

Toolbox

Search