Difference between revisions of "2019 AMC 8 Problems/Problem 15"
(→Solution 1) |
(→Solution 1) |
||
Line 6: | Line 6: | ||
The number of people wearing caps and sunglasses is | The number of people wearing caps and sunglasses is | ||
<math>\frac{2}{5}*35=14</math>. 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. | ||
− | <math>\frac{14}{50}< | + | <math>\frac{14}{50}<cmath>=</cmath>\boxed{\textbf{(B)}\frac{7}{25}}</math>~heeeeeeheeeeee |
==See Also== | ==See Also== |
Revision as of 10:00, 24 November 2019
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 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. ~heeeeeeheeeeee
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.