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

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So, in total the two cases combined equals <math>\frac{1}{27} + \frac{6}{27}</math> = <math>\boxed{\textbf{(D)}\ \frac{7}{27}}</math>.
 
So, in total the two cases combined equals <math>\frac{1}{27} + \frac{6}{27}</math> = <math>\boxed{\textbf{(D)}\ \frac{7}{27}}</math>.
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More detailed explanation: For case 1, the contestant must guess all three correctly. The probability of guessing one problem right is <math>\frac{1}{3}</math>, so the probability of getting all three right is <math>\left(\frac{1}{3}\right)^{3}</math>.
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For case 2: we must choose one of the problems to answer correctly and two to answer incorrectly. The probabilities for guessing correctly and incorrectly are <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math>, respectively. So we have <math>\left(\frac{1}{3}\right)^{2}\cdot\frac{2}{3}\cdot3</math>. The answer is the sum of probabilities of case 1 and 2, since there are no overcounts. <math>\frac{1}{27}+\frac{6}{27}=\frac{7}{27}</math>.
  
 
==Solution 2 (complementary counting)==
 
==Solution 2 (complementary counting)==

Latest revision as of 07:16, 4 November 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 $\frac{1}{27} + \frac{6}{27}$ = $\boxed{\textbf{(D)}\ \frac{7}{27}}$.


More detailed explanation: For case 1, the contestant must guess all three correctly. The probability of guessing one problem right is $\frac{1}{3}$, so the probability of getting all three right is $\left(\frac{1}{3}\right)^{3}$. For case 2: we must choose one of the problems to answer correctly and two to answer incorrectly. The probabilities for guessing correctly and incorrectly are $\frac{1}{3}$ and $\frac{2}{3}$, respectively. So we have $\left(\frac{1}{3}\right)^{2}\cdot\frac{2}{3}\cdot3$. The answer is the sum of probabilities of case 1 and 2, since there are no overcounts. $\frac{1}{27}+\frac{6}{27}=\frac{7}{27}$.

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 getting each question incorrect is $\frac{2}{3}$. Thus, the probability of getting all questions incorrect is $\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}$.

Case 2: The contestant gets one question right. There are 3 ways the contestant can get one question correct 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 $\frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{27}$. Each of these 3 ways has a probability associated with it, so probability of getting 1 correct is $3 * \frac{4}{27} = \frac{4}{9}$

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

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