Difference between revisions of "2019 AMC 8 Problems/Problem 15"
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==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 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? | + | 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? |
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<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> | ||
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==Solution 1== | ==Solution 1== | ||
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. |
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<math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math> | <math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math> | ||
| − | ==Solution Explained== | + | ==Solution 2== |
| + | Let <math>A</math> be the event that a randomly selected person is wearing sunglasses, and let <math>B</math> be the event that a randomly selected person is wearing a cap. We can write <math>P(A \cap B)</math> in two ways: <math>P(A)P(B|A)</math> or <math>P(B)P(A|B)</math>. Suppose there are <math>t</math> people in total. Then <cmath>P(A) = \frac{50}{t}</cmath> and <cmath>P(B) = \frac{35}{t}.</cmath> Additionally, we know that the probability that someone is wearing sunglasses given that they wear a cap is <math>\frac{2}{5}</math>, so <math>P(A|B) = \frac{2}{5}</math>. We let <math>P(B|A)</math>, which is the quantity we want to find, be equal to <math>x</math>. Substituting in, we get <cmath>\frac{50}{t} \cdot x = \frac{35}{t} \cdot \frac{2}{5}</cmath><cmath>\implies 50x = 35 \cdot \frac{2}{5}</cmath> | ||
| + | <cmath>\implies 50x = 14</cmath> | ||
| + | <cmath>\implies x = \frac{14}{50}</cmath> | ||
| + | <cmath>= \boxed{\textbf{(B)}~\frac{7}{25}}</cmath> | ||
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| + | Note: This solution makes use of the dependent events probability formula, <math>P(A \cap B) = P(A)P(B|A)</math>, where <math>P(B|A)</math> represents the probability that <math>B</math> occurs given that <math>A</math> has already occurred and <math>P(A \cup B)</math> represents the probability of both <math>A</math> and <math>B</math> happening. | ||
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| + | ~ cxsmi | ||
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| + | ==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 | https://youtu.be/gOZOCFNXMhE ~ The Learning Royal | ||
| − | ==Video Solution== | + | ===Video Solution by OmegaLearn=== |
| − | https:// | + | https://youtu.be/6xNkyDgIhEE?t=252 |
| − | + | ~ pi_is_3.14 | |
| − | https:// | + | ===Video Solution=== |
| + | https://www.youtube.com/watch?v=gKlYlAiBzrs | ||
| − | == Video Solution == | + | ~ MathEx |
| + | |||
| + | https://www.youtube.com/watch?v=afMsUqER13c | ||
| + | |||
| + | Another video | ||
| + | |||
| + | https://youtu.be/37UWNaltvQo | ||
| + | |||
| + | -Happytwin | ||
| + | |||
| + | === Video Solution === | ||
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=16 | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=16 | ||
| − | ==Video Solution== | + | ===Video Solution=== |
https://youtu.be/9nbaMSAQCNU | https://youtu.be/9nbaMSAQCNU | ||
~savannahsolver | ~savannahsolver | ||
| + | |||
| + | ===Video Solution (MOST EFFICIENT+ CREATIVE THINKING!!!)=== | ||
| + | https://youtu.be/cK_mfkZ_rYM | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ===Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)=== | ||
| + | https://youtu.be/Xm4ZGND9WoY | ||
| + | |||
| + | ~Hayabusa1 | ||
==See Also== | ==See Also== | ||
Latest revision as of 00:28, 6 May 2024
Contents
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?
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(Problem 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. 
