Difference between revisions of "2019 AMC 8 Problems/Problem 15"

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
 
The number of people wearing caps and sunglasses is  
 
The number of people wearing caps and sunglasses is  
<cmath>\left(\frac{2}{5}\right)</cmath>*35=14. So then 14 people out of the 50 people wearing sunglasses also have caps.  
+
<math>\frac{2}{5}*35=14</math>. So then 14 people out of the 50 people wearing sunglasses also have caps.  
<cmath>\left(\frac{14}{50}\right)</cmath>=<math>\boxed{\textbf{(B)}\frac{7}{25}}</math>~heeeeeeheeeeee
+
<math>\frac{14}{50}</math><math>=</math>\boxed{\textbf{(B)}\frac{7}{25}}$~heeeeeeheeeeee
  
 
==See Also==
 
==See Also==

Revision as of 11:00, 24 November 2019

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 $\frac{2}{5}*35=14$. So then 14 people out of the 50 people wearing sunglasses also have caps. $\frac{14}{50}$$=$\boxed{\textbf{(B)}\frac{7}{25}}$~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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