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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.
  
−
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>.
+
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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 +
==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.
 +
 +
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 guessing incorrectly for each question is <math>\frac{2}{3}</math>. Thus, the probability of guessing all questions incorrectly is <math>\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}</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>
  
 
==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}}

Revision as of 20:22, 16 February 2017

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

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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