Difference between revisions of "2019 AMC 8 Problems/Problem 15"
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
| − | ===Video Solution by The Power of Logic( | + | ===Video Solution by The Power of Logic(1 to 25 Full Solution)=== |
https://youtu.be/Xm4ZGND9WoY | https://youtu.be/Xm4ZGND9WoY | ||
Latest revision as of 09:32, 9 November 2024
Contents
Problem
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 2
Let 
be the event that a randomly selected person is wearing sunglasses, and let 
be the event that a randomly selected person is wearing a cap. We can write 
in two ways: 
or 
. Suppose there are 
people in total. Then ![\[P(A) = \frac{50}{t}\]](http://latex.artofproblemsolving.com/d/b/c/dbc343187707bf83976dc6799117d87d9d53c806.png)
and ![\[P(B) = \frac{35}{t}.\]](http://latex.artofproblemsolving.com/1/7/2/172081cf9ae341f6ccd674b57aa6adef57fd3c51.png)
Additionally, we know that the probability that someone is wearing sunglasses given that they wear a cap is , so 
. We let 
, which is the quantity we want to find, be equal to 
. Substituting in, we get ![\[\frac{50}{t} \cdot x = \frac{35}{t} \cdot \frac{2}{5}\]](http://latex.artofproblemsolving.com/7/f/a/7fa08897c48d996f17200ce25b5c6ea135a8c22b.png)
![\[\implies 50x = 35 \cdot \frac{2}{5}\]](http://latex.artofproblemsolving.com/0/4/d/04d370350c482dd77a7b85b169461c7f9742235a.png)
![\[\implies 50x = 14\]](http://latex.artofproblemsolving.com/2/a/1/2a13ba71a39ae3000c4824ac1b38e05e283931fc.png)
![\[\implies x = \frac{14}{50}\]](http://latex.artofproblemsolving.com/1/a/3/1a3432b7fccf475dff21a9b3d18190d7703cf227.png)
![\[= \boxed{\textbf{(B)}~\frac{7}{25}}\]](http://latex.artofproblemsolving.com/c/0/1/c0108c695129183cb29e116e6985d7522b4ebcaf.png)
Note: This solution makes use of the dependent events probability formula, 
, where represents the probability that occurs given that has already occurred and 
represents the probability of both and happening.
~ cxsmi
Video Solutions
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=V_SNrrp17pztxbQG&t=4518
~Math-X
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(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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
