Question

A certain disease has an incidence rate of 0.5%. The false negative rate on a test for the disease is 5%; the false positive rate is 5%. Compute the probability that a person who tests positive actually has the disease. (You may find it useful to construct a probability contingency table.)
The probability a person who tests positive actually has the disease is ________


Give your answer accurate to at least 3 decimal places.

Answer 1

Using conditional probability, it is found that the probability a person who tests positive actually has the disease is 0.087 = 8.7%.

What is Conditional Probability?

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which:

• P(B|A) is the probability of event B happening, given that A happened.

• $P(A \cap B)$ is the probability of both A and B happening.

• P(A) is the probability of A happening.

For this problem, the events are given as follows:

• Event A: Positive test.

• Event B: Person has the disease.

The percentages associated with a positive test is:

• 95% of 0.5%(person has the disease).

• 5% of 99.5%(person does not have the disease).

Hence:

P(A) = 0.95 x 0.005 + 0.05 x 0.995 = 0.0545.

The probability of both a positive test and having the disease is:

$P(A \cap B) = 0.95 \times 0.005 = 0.00475$

Hence the conditional probability is:

P(B|A) = 0.00475/0.0545 = 0.087.

The probability a person who tests positive actually has the disease is 0.087 = 8.7%.

More can be learned about conditional probability at brainly.com/question/14398287

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