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Difference between revisions of "2012 AMC 10B Problems/Problem 18"

(Created page with "==Problem 18== 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...")
 
(→‎Solution 18)
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This question can be solved by considering all the possibilities:
 
This question can be solved by considering all the possibilities:
  
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<math>1</math> out of <math>500</math> will have the disease and will be tested positive for the disease.
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<math>1</math> out of <math>500</math> people will have the disease and will be tested positive for the disease.
  
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Out of the remaining <math>499</math>, <math>2\%</math>, or <math>9.98</math> people, will be tested positive for the disease incorrectly.
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Out of the remaining <math>499</math> people, <math>2\%</math>, or <math>9.98</math> people, will be tested positive for the disease incorrectly.
  
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Therefore, <math>p</math> can be found by taking <math>\dfrac{1}{1+9.98}</math>, whichc is closest to <math>\dfrac{1}{11}</math>,or <math>\textbf{(C)}}</math>
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Therefore, <math>p</math> can be found by taking <math>\dfrac{1}{1+9.98}</math>, which is closest to <math>\dfrac{1}{11}</math>,or <math>\textbf{(C)}}</math>

Revision as of 08:33, 27 February 2012

Problem 18

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 18

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 $\dfrac{1}{11}$,or $\textbf{(C)}}$ (Error compiling LaTeX. Unknown error_msg)

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