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

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The number of people wearing caps and sunglasses is  
 
The number of people wearing caps and sunglasses is  
 
<math>\frac{2}{5}\cdot35=14</math>. So then 14 people out of the 50 people wearing sunglasses also have caps.  
 
<math>\frac{2}{5}\cdot35=14</math>. So then 14 people out of the 50 people wearing sunglasses also have caps.  
<math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math>~heeeeeeheeeeee
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<math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:48, 17 May 2020

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 $\frac{2}{5}\cdot35=14$. So then 14 people out of the 50 people wearing sunglasses also have caps. $\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}$

See Also

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

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