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 | + | 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> | ||
==Solution 1== | ==Solution 1== | ||
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> | <math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math> | ||
+ | |||
+ | ==Solution Explained== | ||
+ | https://youtu.be/gOZOCFNXMhE ~ The Learning Royal | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/6xNkyDgIhEE?t=252 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=gKlYlAiBzrs | ||
+ | |||
+ | ~ MathEx | ||
+ | |||
+ | https://www.youtube.com/watch?v=afMsUqER13c | ||
+ | |||
+ | Another video | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/9nbaMSAQCNU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution (MOST EFFICIENT+ CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/cK_mfkZ_rYM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
==See Also== | ==See Also== |
Latest revision as of 21:27, 26 October 2023
Contents
Problem 15
On a beach people are wearing sunglasses and 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 . If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?
Solution 1
The number of people wearing caps and sunglasses is . So then, 14 people out of the 50 people wearing sunglasses also have caps.
Solution Explained
https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=252
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=gKlYlAiBzrs
~ MathEx
https://www.youtube.com/watch?v=afMsUqER13c
Another video
-Happytwin
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=16
Video Solution
~savannahsolver
Video Solution (MOST EFFICIENT+ CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
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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