Difference between revisions of "2019 AMC 8 Problems/Problem 15"
Mathboy282 (talk | contribs) m (→Solution 1) |
m (→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}\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> |
==See Also== | ==See Also== | ||
Revision as of 19:48, 17 May 2020
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.
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. 
