2012 AMC 10B Problem 18

Question

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?

\dfrac{1}{98}

\dfrac{1}{9}

\dfrac{1}{11}

\dfrac{49}{99}

\dfrac{98}{99}

Answer: C

Concepts:conditional probability Bayes’ Theorem

Difficulty rating: 1730

Small hint:

Out of

500

people, about

1

true positive and

10

false positives occur

Big hint:

Condition on the test being positive

Solution:

Among

500

people, about

1

person has the disease and tests positive. Of the remaining

499

people, about

2\%

test falsely positive, which is about

10

people.

So among about

11

positive tests, only about

1

is a true positive. The probability is closest to

\dfrac1{11}

.

Thus, C is the correct answer.

2012 AMC 10B Problem 18

Answer: C

Difficulty rating: 1730

Small hint:

Big hint:

Solution:

Problem 18 in other years