AMC 10 · 2012 · #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

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 $p$ ?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: $1$ out of every $500$ people has the disease; A person with the disease always tests positive; A person without the disease tests positive $2\%$ of the time (false positive); Answer choices: (A) $\frac{1}{98}$, (B) $\frac{1}{9}$, (C) $\frac{1}{11}$, (D) $\frac{49}{99}$, (E) $\frac{98}{99}$

Unknowns: The probability $p$ that a person who tests positive truly has the disease

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

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.

Execute — Answer: C

#9 Solve an Easier Related Problem 7.SP.C.7 Step 1

Pick a concrete crowd

Rather than track probabilities, imagine a specific population.
Choose $50{,}000$ people, a multiple of $500$, so that "$1$ in $500$" and "$2\%$" both land on whole numbers.

$$N = 50{,}000 \text{ people}$$

💡 A fixed crowd size turns fuzzy chances into countable people.

#7 Identify Subproblems 7.RP.A.3 Step 2

Count the true positives

With $1$ sick person in every $500$, the number who are sick is $50{,}000 \div 500 = 100$.
Every one of them tests positive, so there are $100$ true positives.

$$\dfrac{50{,}000}{500} = 100 \text{ true positives}$$

💡 Sick people never slip past the test, so their count is exactly the positives you want.

#7 Identify Subproblems 7.RP.A.3 Step 3

Count the false positives

The remaining $50{,}000 - 100 = 49{,}900$ people are healthy, and $2\%$ of them test positive by mistake.
That gives $0.02 \times 49{,}900 = 998$ false positives.

$$0.02 \times 49{,}900 = 998 \text{ false positives}$$

💡 A tiny error rate still snags many people because the healthy group is huge.

#7 Identify Subproblems 6.RP.A.3 Step 4

Form the fraction of positives who are sick

Everyone who tests positive is either truly sick or falsely flagged: $100 + 998 = 1098$ positives in all.
The share who are genuinely sick is $p = \dfrac{100}{1098}$.

$$p = \dfrac{100}{100 + 998} = \dfrac{100}{1098} \approx 0.091$$

💡 $p$ is just the sick slice measured against the entire positive pile.

#3 Eliminate Possibilities 6.RP.A.3 Step 5

Match to the closest choice

Now compare $0.091$ to the options.
Since $\frac{1}{11} \approx 0.0909$ while $\frac{1}{9} \approx 0.111$ and $\frac{1}{98} \approx 0.010$, the closest value is $\frac{1}{11}$, so the answer is (C).

$$\dfrac{100}{1098} \approx 0.091 \approx \dfrac{1}{11} \;\Rightarrow\; \textbf{(C)}$$

💡 Turning the fraction into a decimal makes the nearest listed value easy to spot.

[1] #9 7.SP.C.7 Rather than track probabilities, imagine a specific population. Choose $50{,}000

[2] #7 7.RP.A.3 With $1$ sick person in every $500$, the number who are sick is $50{,}000 \div 5

[3] #7 7.RP.A.3 The remaining $50{,}000 - 100 = 49{,}900$ people are healthy, and $2\%$ of them

[4] #7 6.RP.A.3 Everyone who tests positive is either truly sick or falsely flagged: $100 + 998

[5] #3 6.RP.A.3 Now compare $0.091$ to the options. Since $\frac{1}{11} \approx 0.0909$ while $\

Review

Reasonableness: 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) $\frac{49}{99}$ and (E) $\frac{98}{99}$. The value $0.091$ is much closer to $\frac{1}{11} \approx 0.0909$ than to $\frac{1}{9} \approx 0.111$, confirming (C).

Alternative: Work in one clean batch of $500$ people. Exactly $1$ is sick and tests positive; the other $499$ are healthy and give $0.02 \times 499 = 9.98$ false positives. Then $p = \dfrac{1}{1 + 9.98} = \dfrac{1}{10.98} \approx 0.091$, which is again closest to $\frac{1}{11}$, matching (C).

CCSS standards used (min grade 7)

7.SP.C.7 Develop probability models and use them to find probabilities of events (Modeling the situation with a concrete population of $50{,}000$ so each stated chance becomes a countable group of people.)
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems (Finding the $100$ sick people from "$1$ in $500$" and the $998$ false positives from "$2\%$ of $49{,}900$.")
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Forming $p = \frac{100}{1098}$ and comparing it to the answer choices as a decimal.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

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

Review

CCSS standards used (min grade 7)

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