Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution by TheBeautyofMath)
(→‎Solution 2 (complementary counting))
Line 25: Line 25:
 
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 <math>\frac{1}{3}</math> so the probability of guessing one correctly and two incorrectly is <math>3* \frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{9}</math>.  
 
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 <math>\frac{1}{3}</math> so the probability of guessing one correctly and two incorrectly is <math>3* \frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{9}</math>.  
  
−
The sum of the two cases is <math>\frac{8}{27} + \frac{4}{9} = \frac{20}{27}</math>. This is the complement of what we want to the answer is <math>1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}</math>  
+
The sum of the two cases is <math>\frac{8}{27} + \frac{4}{9} = \frac{20}{27}</math>. This is the complement of what we want so the answer is <math>1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:23, 25 July 2021

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 so the answer is $1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}$

Video Solution

https://youtu.be/XYeexmAyVzQ

~savannahsolver

Video Solution by TheBeautyofMath

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

~IceMatrix

Video Solution

https://youtu.be/IRyWOZQMTV8?t=1029

~ pi_is_3.14

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_9&oldid=159117"