2019 AMC 8 Problems/Problem 15

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.

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Video Solution by EzLx

https://youtu.be/SkDf39Cg5z4

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Video Solution

https://youtu.be/6xNkyDgIhEE?t=250pih-jsm

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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