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Difference between revisions of "2017 AMC 12B Problems/Problem 10"

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WLOG, let there be <math>100</math> students. <math>60</math> of them like dancing, and <math>40</math> do not. Of those who like dancing, <math>20\%</math>, or <math>12</math> of them say they dislike dancing. Of those who dislike dancing, <math>90\%</math>, or <math>36</math> of them say they dislike it. Thus, <math>\frac{12}{12+36} = \frac{12}{48} = \frac{1}{4} = 25\% \boxed{\textbf{(D)}}</math>
 
WLOG, let there be <math>100</math> students. <math>60</math> of them like dancing, and <math>40</math> do not. Of those who like dancing, <math>20\%</math>, or <math>12</math> of them say they dislike dancing. Of those who dislike dancing, <math>90\%</math>, or <math>36</math> of them say they dislike it. Thus, <math>\frac{12}{12+36} = \frac{12}{48} = \frac{1}{4} = 25\% \boxed{\textbf{(D)}}</math>
  
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==Solution 2- Bayes' Theorem==
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==Solution 2 (Bayes' Theorem)==
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The question can be translated into P{Likes|Says Dislike}).  
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The question can be translated into P(Likes|Says Dislike).  
  
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This is equal to the probability of <math>\frac{\textnormal{P(Likes} \cap \textnormal{Says Dislike)}}{\textnormal{P(Says Dislike)}}</math>. P(Likes <math>\cap</math> Says Dislike) = 0.6 <math>\cdot</math> 0.2 = 0.12. P(Says Dislike) = (0.4 <math>\cdot</math> 0.9) + (0.6 <math>\cdot</math> 0.2) = 0.48. 0.12/0.48 = 25%
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By Bayes' Theorem, this is equal to the probability of <math>\frac{\textnormal{P(Likes} \cap \textnormal{Says Dislike)}}{\textnormal{P(Says Dislike)}}</math>. <math>\textnormal{P(Likes} \cap \textnormal{Says Dislike)} = 0.6 \cdot 0.2 = 0.12</math>, and <math>\textnormal{P(Says Dislike)} = (0.4 \cdot 0.9) + (0.6 \cdot 0.2) = 0.48</math>.
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Therefore, you get a probability of <math>\frac{0.12}{0.48} = 25\% \boxed{\textbf{(D)}}</math>
  
 
~Directrixxx
 
~Directrixxx

Latest revision as of 06:09, 25 October 2024

Problem

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ \frac{100}{3}\%$

Solution 1

WLOG, let there be $100$ students. $60$ of them like dancing, and $40$ do not. Of those who like dancing, $20\%$, or $12$ of them say they dislike dancing. Of those who dislike dancing, $90\%$, or $36$ of them say they dislike it. Thus, $\frac{12}{12+36} = \frac{12}{48} = \frac{1}{4} = 25\% \boxed{\textbf{(D)}}$

Solution 2 (Bayes' Theorem)

The question can be translated into P(Likes|Says Dislike).

By Bayes' Theorem, this is equal to the probability of $\frac{\textnormal{P(Likes} \cap \textnormal{Says Dislike)}}{\textnormal{P(Says Dislike)}}$. $\textnormal{P(Likes} \cap \textnormal{Says Dislike)} = 0.6 \cdot 0.2 = 0.12$, and $\textnormal{P(Says Dislike)} = (0.4 \cdot 0.9) + (0.6 \cdot 0.2) = 0.48$. Therefore, you get a probability of $\frac{0.12}{0.48} = 25\% \boxed{\textbf{(D)}}$

~Directrixxx

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 12 Problems and Solutions

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