2012 AMC 10B Problem 18

P16-20~4 minPrint

Conditional Probability & States Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

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 $p$ ?

Show solution without answering

Key insight

Imagine 50,000 people: 100 are sick and test positive, but 2% of the 49,900 healthy people, 998 people, also test positive. So about 1 in 11 positives is real.

Solution

Probability questions like this are easiest if you imagine an actual crowd of people and count.

1. Pick a crowd size that makes the numbers whole. Take $50{,}000$ people.
2. One in every $500$ has the disease: $50{,}000 \div 500 = 100$ sick people. They all test positive.
3. The other $50{,}000 - 100 = 49{,}900$ people are healthy. Of them, $2\%$ test positive by mistake: $0.02 \times 49{,}900 = 998$ people.
4. So the people with a positive test are $100 + 998 = 1098$ in all.
5. We want the chance that a person with a positive test is really sick. That is the sick positives out of all positives:
$p = \frac{100}{1098} \approx \frac{100}{1100} = \frac{1}{11}.$
6. Compare with the choices: $\frac1{11} \approx 0.091$ and $\frac19 \approx 0.111$ . The value $\frac{100}{1098} \approx 0.091$ is closest to $\frac1{11}$ .

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

Why this works

The question only looks at people who tested positive and asks what fraction of them are sick. Healthy people far outnumber sick people, so even a small $2\%$ error rate among the healthy produces about ten times as many false alarms as real cases. Counting an imaginary crowd turns a confusing probability question into simple division.

Alternative approach

Use a crowd of $500$ instead. Exactly $1$ person is sick, and about $0.02 \times 499 = 9.98$ healthy people test positive. Then $p = \frac{1}{1 + 9.98} = \frac{1}{10.98}$ , which is very close to $\frac{1}{11}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Imagine testing a group of $500$ people.

Exactly $1$ person has the disease, and that person tests positive.
The other $499$ are healthy, and $2\%$ of them test positive anyway: $0.02 \cdot 499 = 9.98$ false positives.

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

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Introduction to Algebra, Chapter 6: Ratios and Percents
Intermediate Counting & Probability, Chapter 11: Conditional Probability
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)