2012 AMC 10B Problems/Problem 18

Problem

Suppose that one of every 500 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 $2\%$ false positive rate--in other words, for such people, $98\%$ of the time the test will turn out negative, but $2\%$ of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let $p$ 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 $p$ ?

$\textbf{(A)}\ \frac{1}{98}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{1}{11}\qquad\textbf{(D)}\ \frac{49}{99}\qquad\textbf{(E)}\ \frac{98}{99}$

Solution

This question can be solved by considering all the possibilities:

$1$ out of $500$ people will have the disease and will be tested positive for the disease.

Out of the remaining $499$ people, $2\%$ , or $9.98$ people, will be tested positive for the disease incorrectly.

Therefore, $p$ can be found by taking $\dfrac{1}{1+9.98}$ , which is closest to $\dfrac{1}{1+10}$ or $\dfrac{1}{11}$ ,or $\boxed{\textbf{(C)}}$

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

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2012 AMC 10B Problems/Problem 18

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