# 2017 AMC 12B Problems/Problem 10

## 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). 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

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