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

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==Problem 18==
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==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 <math>2\%</math> false positive rate--in other words, for such people, <math>98\%</math> of the time the test will turn out negative, but <math>2\%</math> of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let <math>p</math> 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 <math>p</math>?
 
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 <math>2\%</math> false positive rate--in other words, for such people, <math>98\%</math> of the time the test will turn out negative, but <math>2\%</math> of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let <math>p</math> 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 <math>p</math>?
  
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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.
 
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>, which is closest to <math>\dfrac{1}{11}</math>,or <math>\boxed{\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}{1+10}</math> or <math>\dfrac{1}{11}</math>,or <math>\boxed{\textbf{(C)}}</math>
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==Solution 2==
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This question can also be solved by using a <math>\dfrac{probability1}{probability2}</math> solution.
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If the test is positive there are two cases: (1) The person was true-positive (2) The person was false-positive
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For the first case, the probability of the person being true-positive is <math>\dfrac{1}{500}</math>.
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For the second case, the probability of the person being false-positive is <math>\dfrac{499}{500}\cdot\dfrac{1}{50}</math>
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The probability of the test being positive is the sum of these values, <math>\dfrac{549}{25000}</math>
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The final probability will be in the form: <math>\dfrac{P(true positive)}{P(positive)}</math>.
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Plugging in the values we get: <math>\dfrac{\dfrac{1}{500}}{\dfrac{549}{25000}}=\dfrac{50}{549}</math> This fraction is clearly closest to  <math>\dfrac{1}{11}</math>, or <math>\boxed{\textbf{(C)}}</math>
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==Solution 3==
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To solve this problem, you need to find the probability of a correctly identified positive and divided by the total probability of a positive.
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The probability of obtaining a correctly identified positive is <math>\dfrac{1}{500}</math>.
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The probability of obtaining an incorrectly identified positive is <math>\dfrac{2}{100}</math> = <math>\dfrac{10}{500}</math>.
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<math>\dfrac{1}{500}</math> + <math>\dfrac{10}{500}</math> = <math>\dfrac{11}{500}</math> = the probability of obtaining a positive.
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Correctly identifed positive / positive = <math>\dfrac{1}{500}</math> / <math>\dfrac{11}{500}</math> = <math>\dfrac{1}{11}</math> or <math>\boxed{\textbf{(C)}}</math>
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==Video Solution==
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https://youtu.be/igGLCogR-dk
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~savannahsolver
  
 
== See Also ==
 
== See Also ==
  
  
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{{AMC10 box|year=2012|ab=B|num-b=17|num-a=20}}
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{{AMC10 box|year=2012|ab=B|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 11:52, 26 September 2024

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 $\boxed{\textbf{(C)}}$

Solution 2

This question can also be solved by using a $\dfrac{probability1}{probability2}$ solution.

If the test is positive there are two cases: (1) The person was true-positive (2) The person was false-positive

For the first case, the probability of the person being true-positive is $\dfrac{1}{500}$.

For the second case, the probability of the person being false-positive is $\dfrac{499}{500}\cdot\dfrac{1}{50}$

The probability of the test being positive is the sum of these values, $\dfrac{549}{25000}$

The final probability will be in the form: $\dfrac{P(true positive)}{P(positive)}$.

Plugging in the values we get: $\dfrac{\dfrac{1}{500}}{\dfrac{549}{25000}}=\dfrac{50}{549}$ This fraction is clearly closest to $\dfrac{1}{11}$, or $\boxed{\textbf{(C)}}$

Solution 3

To solve this problem, you need to find the probability of a correctly identified positive and divided by the total probability of a positive.

The probability of obtaining a correctly identified positive is $\dfrac{1}{500}$.

The probability of obtaining an incorrectly identified positive is $\dfrac{2}{100}$ = $\dfrac{10}{500}$.

$\dfrac{1}{500}$ + $\dfrac{10}{500}$ = $\dfrac{11}{500}$ = the probability of obtaining a positive.

Correctly identifed positive / positive = $\dfrac{1}{500}$ / $\dfrac{11}{500}$ = $\dfrac{1}{11}$ or $\boxed{\textbf{(C)}}$

Video Solution

https://youtu.be/igGLCogR-dk

~savannahsolver

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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