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

 
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==Problem 15==
 
==Problem 15==
On a beach <math>50</math> people are wearing sunglasses and <math>35</math> 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 <math>\frac{2}{5}</math>. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?
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On a beach <math>50</math> people are wearing sunglasses and <math>35</math> 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 also wearing sunglasses is <math>\frac{2}{5}</math>. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?
 
<math>\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}</math>
 
<math>\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}</math>
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==Video Solution==
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https://youtu.be/6xNkyDgIhEE?t=252
  
 
==Solution 1==
 
==Solution 1==
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<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>
 
<math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math>
 
  
 
==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx
 
https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx
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Another video - https://www.youtube.com/watch?v=afMsUqER13c
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https://youtu.be/37UWNaltvQo -Happytwin
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== Video Solution ==
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=16
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2019|num-b=14|num-a=16}}
 
{{AMC8 box|year=2019|num-b=14|num-a=16}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:32, 23 April 2021

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 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? $\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$

Video Solution

https://youtu.be/6xNkyDgIhEE?t=252

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. $\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}$

Video Solution

https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx

Another video - https://www.youtube.com/watch?v=afMsUqER13c

https://youtu.be/37UWNaltvQo -Happytwin

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=16

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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