2019 AMC 8 Problems/Problem 15

Revision as of 14:19, 20 November 2019 by Heeeeeeeheeeee (talk | contribs) (Solution 1)

Problem 15

On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is $\frac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap? $\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$

Solution 1

The number of people wearing caps and sunglasses is \[\left(\frac{2}{5}\right)\]*35=14. So then 14 people out of the 50 people wearing sunglasses also have caps. \[\left(\frac{14}{50}\right)\]=$\boxed{\textbf{(B)}\frac{7}{25}}$~heeeeeeheeeeee

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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