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2012 AMC 10B Problem 18

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2012 AMC 10B Problem 18AMC 10 Step by Step
Name:                             Date: September 13, 2026
Conditional Probability & States · difficulty 3 of 5 · about 4 min · position P16-20

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%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 pp ?

  1. A)

    198\frac{1}{98}

  2. B)

    19\frac{1}{9}

  3. C)

    111\frac{1}{11}

  4. D)

    4999\frac{49}{99}

  5. E)

    9899\frac{98}{99}

Answer

C

Key insight

Picture 500 people: 1 true positive and about 0.02 * 499 ≈ 10 false positives, so roughly 1 in 11 positive tests is real.

Solution

Imagine testing a group of 500500 people.

  • Exactly 11 person has the disease, and that person tests positive.
  • The other 499499 are healthy, and 2%2\% of them test positive anyway: 0.02⋅499=9.980.02 \cdot 499 = 9.98 false positives.

So about 1+9.98=10.981 + 9.98 = 10.98 people test positive, and only 11 of them is actually sick:
p=110.98≈0.091.p = \frac{1}{10.98} \approx 0.091.
Among the choices, 111≈0.0909\frac{1}{11} \approx 0.0909 is closest (19≈0.111\frac19 \approx 0.111 is noticeably farther).

The answer is (C) 111\boxed{\textbf{(C)}\ \frac{1}{11}}.

Why this works

"Probability of disease given a positive test" restricts attention to the positive testers and asks what fraction of them are sick. Because healthy people vastly outnumber sick ones, even a small false-positive rate produces more false positives than true positives, which is why the answer is small. Choosing a concrete population (here 500500, or 50,00050{,}000 to avoid decimals) makes Bayes' rule a matter of counting people.

Alternative approach

Formally, p=1500⋅11500⋅1+499500⋅0.02=11+9.98=110.98p = \dfrac{\frac{1}{500}\cdot 1}{\frac{1}{500}\cdot 1 + \frac{499}{500}\cdot 0.02} = \dfrac{1}{1 + 9.98} = \dfrac{1}{10.98}. With 50,00050{,}000 people the counts are whole: 100100 true positives and 998998 false positives, so p=1001098≈111p = \frac{100}{1098} \approx \frac{1}{11}.

The trap

Answering 98/99 or 49/99 by confusing 'positive given disease' with 'disease given positive', ignoring how rare the disease is.

Common mistakes

  • Answering 98/99 or 49/99 by confusing 'positive given disease' with 'disease given positive', ignoring how rare the disease is.
  • Rounding 499499 to 500500 is fine, but then choosing 19\frac{1}{9} instead of 111\frac{1}{11} because 1010 false positives plus 11 true positive was miscounted as 99.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer