Difference between revisions of "2017 AMC 12B Problems/Problem 10"
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<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ \frac{100}{3}\%</math> | <math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ \frac{100}{3}\%</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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> | ||
| + | |||
| + | ==Solution 2 (Bayes' Theorem)== | ||
| + | The question can be translated into P(Likes|Says Dislike). | ||
| + | |||
| + | 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>. | ||
| + | Therefore, you get a probability of <math>\frac{0.12}{0.48} = 25\% \boxed{\textbf{(D)}}</math> | ||
| + | |||
| + | ~Directrixxx | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2017|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 06:09, 25 October 2024
Problem
At Typico High School, 
of the students like dancing, and the rest dislike it. Of those who like dancing, 
say that they like it, and the rest say that they dislike it. Of those who dislike dancing, 
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?
Solution 1
WLOG, let there be 
students. 
of them like dancing, and 
do not. Of those who like dancing, 
, or 
of them say they dislike dancing. Of those who dislike dancing, , or 
of them say they dislike it. Thus, 
Solution 2 (Bayes' Theorem)
The question can be translated into P(Likes|Says Dislike).
By Bayes' Theorem, this is equal to the probability of 
. 
, and 
.
Therefore, you get a probability of 
~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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