2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 15 Medium

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?

  • A.

    \frac{14}{85}

  • B.

    \frac{7}{25}

  • C.

    \frac{2}{5}

  • D.

    \frac{4}{7}

  • E.

    \frac{7}{10}

Answer:B

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}}

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