Difference between revisions of "2019 AMC 8 Problems/Problem 15"
Sum-of-sums (talk | contribs) m (→Video Solution) |
m (→Problem 15) |
||
Line 1: | Line 1: | ||
==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> | ||
Revision as of 16:14, 1 February 2021
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?
Video Solution
https://youtu.be/6xNkyDgIhEE?t=252
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.
Video Solution
https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx
Another video - https://www.youtube.com/watch?v=afMsUqER13c
https://youtu.be/37UWNaltvQo -Happytwin
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.