2012 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2012 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10B solutions, or check the answer key.

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Concepts:conditional probabilityBayes’ Theorem

Difficulty rating: 1730

18.

Suppose that one of every 500500 people in a certain population has a particular disease, which displays no symptoms. A blood test is available for screening for this disease. For a person who has this disease, the test always turns out positive.

For a person who does not have the disease, however, there is a 2%2\% false positive rate. In other words, for such people, 98%98\% of the time the test will turn out negative, but 2%2\% of the time the test will turn out positive and will incorrectly indicate that the person has the disease.

Let pp be the probability that a person who is chosen at random from this population and gets a positive test result actually has the disease. Which of the following is closest to p?p?

198 \dfrac{1}{98}

19 \dfrac{1}{9}

111 \dfrac{1}{11}

4999 \dfrac{49}{99}

9899 \dfrac{98}{99}

Solution:

Among 500500 people, about 11 person has the disease and tests positive. Of the remaining 499499 people, about 2%2\% test falsely positive, which is about 1010 people.

So among about 1111 positive tests, only about 11 is a true positive. The probability is closest to 111\dfrac1{11}.

Thus, C is the correct answer.

Problem 18 in Other Years