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

m (→‎Solution 2- Bayes' Theorem)
(→‎Solution 2- Bayes' Theorem)
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==Solution 2- Bayes' Theorem==
 
==Solution 2- Bayes' Theorem==
−
The question can be translated into <math>P(\textnormal {Likes|Says Dislike})</math>.  
+
The question can be translated into <math>P(\textnormal {Likes \middle Says Dislike})</math>.  
  
 
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) = .6 <math>\cdot</math> .2 = .12. P(Says Dislike) = (.4 <math>\cdot</math> .9) + (.6 <math>\cdot</math> .2) = .48. .12/.48 = 25%
 
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) = .6 <math>\cdot</math> .2 = .12. P(Says Dislike) = (.4 <math>\cdot</math> .9) + (.6 <math>\cdot</math> .2) = .48. .12/.48 = 25%

Revision as of 11:58, 3 July 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 $P(\textnormal {Likes \middle Says Dislike})$ (Error compiling LaTeX. Unknown error_msg).

This is equal to the probability of $\frac{\textnormal{P(Likes} \cap \textnormal{Says Dislike)}}{\textnormal{P(Says Dislike)}}$. P(Likes $\cap$ Says Dislike) = .6 $\cdot$ .2 = .12. P(Says Dislike) = (.4 $\cdot$ .9) + (.6 $\cdot$ .2) = .48. .12/.48 = 25%

~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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