2017 AMC 12B Problems/Problem 10

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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 (ProblemsAnswer KeyResources)
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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