Difference between revisions of "2019 AMC 8 Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
| + | The number of people wearing caps and sunglasses is | ||
| + | <cmath>\left(\frac{2}{5}\right)</cmath>*35=14. So then 14 people out of the 50 people wearing sunglasses also have caps. | ||
| + | <cmath>\left(\frac{14}{50}\right)</cmath>=<math>\boxed{\textbf{(B)}\frac{7}{25}}</math>~heeeeeeheeeeee | ||
==See Also== | ==See Also== | ||
Revision as of 14:19, 20 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
![\[\left(\frac{2}{5}\right)\]](http://latex.artofproblemsolving.com/f/5/f/f5f46019d93e93e727efa57eec86e9c0bf07b95f.png)
*35=14. So then 14 people out of the 50 people wearing sunglasses also have caps.
![\[\left(\frac{14}{50}\right)\]](http://latex.artofproblemsolving.com/3/6/1/36181534f59864ee29ae4369b38c4a9730899d51.png)
=
~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 | ||
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