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Difference between revisions of "2019 AMC 8 Problems/Problem 15"
 
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==Problem 15==
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==Problem==
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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  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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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?  
 +
 
 
<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>
  
 
==Solution 1==
 
==Solution 1==
 
The number of people wearing caps and sunglasses is  
 
The number of people wearing caps and sunglasses is  
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<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{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>
  
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==Solution 2==
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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>
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<cmath>\implies 50x = 14</cmath>
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<cmath>\implies x = \frac{14}{50}</cmath>
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<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.
 +
 +
~ 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
 +
 +
===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
 +
 +
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
 +
 +
===Video Solution===
 +
https://youtu.be/9nbaMSAQCNU
 +
 +
~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(1 to 25 Full Solution)===
 +
https://youtu.be/Xm4ZGND9WoY
 +
 +
~Hayabusa1
  
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==Video Solution==
 
−
https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx
 
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2019|num-b=14|num-a=16}}
 
{{AMC8 box|year=2019|num-b=14|num-a=16}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:32, 9 November 2024

Problem

On a beach $50$ people are wearing sunglasses and $35$ 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 $\frac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?

$\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$

Solution 1

The number of people wearing caps and sunglasses is $\frac{2}{5}\cdot35=14$. So then, 14 people out of the 50 people wearing sunglasses also have caps.

$\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}$

Solution 2

Let $A$ be the event that a randomly selected person is wearing sunglasses, and let $B$ be the event that a randomly selected person is wearing a cap. We can write $P(A \cap B)$ in two ways: $P(A)P(B|A)$ or $P(B)P(A|B)$. Suppose there are $t$ people in total. Then \[P(A) = \frac{50}{t}\] and \[P(B) = \frac{35}{t}.\] Additionally, we know that the probability that someone is wearing sunglasses given that they wear a cap is $\frac{2}{5}$, so $P(A|B) = \frac{2}{5}$. We let $P(B|A)$, which is the quantity we want to find, be equal to $x$. Substituting in, we get \[\frac{50}{t} \cdot x = \frac{35}{t} \cdot \frac{2}{5}\]\[\implies 50x = 35 \cdot \frac{2}{5}\] \[\implies 50x = 14\] \[\implies x = \frac{14}{50}\] \[= \boxed{\textbf{(B)}~\frac{7}{25}}\]

Note: This solution makes use of the dependent events probability formula, $P(A \cap B) = P(A)P(B|A)$, where $P(B|A)$ represents the probability that $B$ occurs given that $A$ has already occurred and $P(A \cup B)$ represents the probability of both $A$ and $B$ 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

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

Video Solution

https://youtu.be/9nbaMSAQCNU

~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(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~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

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