Competition · AMC preparation · step 4 of 4

AMC 10 · 2012B · #18

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype Two-Attribute Table: Fill or Optimize a Cell

conditional-probabilityprobability-basic easier-related-problem ↑ Prerequisites: probability-basic

📏 Medium solution 💡 3 insights

Problem

In a population, 1 of every 500 people has a silent disease. The test is always positive for someone who has the disease, and it is positive 2% of the time for someone who does not. Among people who test positive, find the fraction who actually have the disease, and choose the answer closest to it.

Pick an answer.

(A)

$\frac{1}{98}$

(B)

$\frac{1}{9}$

(C)

$\frac{1}{11}$

(D)

$\frac{49}{99}$

(E)

$\frac{98}{99}$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Percents and "1 in 500" are hard to juggle as bare probabilities, so Tool #9 (Solve an Easier Related Problem) replaces them with a concrete crowd of 50,000 people, turning every chance into a plain head count. Tool #7 (Identify Subproblems) splits the positives into the two ways they happen — truly sick versus healthy-but-flagged — and counts each separately. Tool #3 (Eliminate Possibilities) is the finish: the exact fraction is compared against the five listed values to pick the closest.

1STEP 1

Pick a concrete crowd

Rather than juggle probabilities, imagine a crowd of 50,000 people — a multiple of 500, so every count lands on a whole number.

N = 50,000 people

2STEP 2

Count the true positives

One sick person in every 500 gives 50,000 ÷ 500 = 100 sick people, and every one of them tests positive.

50,000/500 = 100 true positives

3STEP 3

Count the false positives

The other 49,900 people are healthy, and 2% of them slip through: 0.02 × 49,900 = 998 false positives.

0.02 × 49,900 = 998 false positives

4STEP 4

Form the fraction of positives who are sick

Every positive is truly sick or falsely flagged, so 100 + 998 = 1098 test positive and the sick share is $\frac{100}{1098}$ .

p = 100/(100 + 998) = 100/1098 ≈ 0.091

5STEP 5

Match to the closest choice

That share is about 0.091, and $\frac{1}{11}$ ≈ 0.0909 sits far closer than $\frac{1}{9}$ ≈ 0.111, so the answer is (C).

100/1098 ≈ 0.091 ≈ 1/11 → (C)

Answer

1/11

The result feels right: even though the test never misses a sick person, sick people are so rare (1 in 500) that the small 2% error rate on a large healthy group produces almost ten times as many false alarms as true cases. So a positive result should mean only about a 1-in-11 chance of really being sick — far below a coin flip, which rules out (D) 49/99 and (E) 98/99. The value 0.091 is much closer to 1/11 ≈ 0.0909 than to 1/9 ≈ 0.111, confirming (C).

💡Key takeaway

Turn the chances into a real crowd, count the true and false positives separately, and p is just the true positives over all the positives.

Pick a concrete crowd
Count the true positives
Count the false positives
Form the fraction of positives who are sick
Match to the closest choice

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Pick a concrete crowd

Count the true positives

Count the false positives

Form the fraction of positives who are sick

Match to the closest choice

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