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

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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution==
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==Solution 1==
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There are two ways that the contestant can win.
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There are two ways the contestant can win.
  
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Case 1: They guess all three right. This can only happen <math>\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}</math> of the time.
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Case 1: The contestant guesses all three right. This can only happen <math>\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}</math> of the time.
  
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Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>\frac{1}{3} * \frac{1}{3} * \frac{2}/3</math> of the time. Thus, <math>\frac{2}{27} * \frac{3}{1}</math> = <math>\frac{6}{27}</math>.
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Case 2: The contestant guesses only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>\frac{1}{3} * \frac{1}{3} * \frac{2}{3}</math> of the time. Thus, <math>\frac{2}{27} * 3</math> = <math>\frac{6}{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>.
 
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>.
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==Solution 2 (complementary counting)==
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Complementary counting is good for solving the problem and checking work if you solved it using the method above.
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There are two ways the contestant can lose.
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Case 1: The contestant guesses zero questions correctly.
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 +
The probability of getting each question incorrect is <math>\frac{2}{3}</math>. Thus, the probability of getting all questions incorrect is <math>\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}</math>.
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 +
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 <math>\frac{1}{3}</math> so the probability of guessing one correctly and two incorrectly is <math> \frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{27}</math>. Each of these 3 ways has a probability associated with it, so probability of getting 1 correct is <math>3 * \frac{4}{27} = \frac{4}{9} </math>
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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>
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==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==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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)
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Problem 8 		Followed by
Problem 10
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All AMC 10 Problems and Solutions

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