Question

A certain disease has an incident rate of 0.5%. If the false negative rate is 4% and the false positive rate is 1%, compute the probability that a person who tests positive actually has it.

Answer 1

To compute the probability that a person who tests positive actually has the disease, we can use Bayes' theorem. The probability is approximately 0.3311, or 33.11%.

Step-by-step explanation:

To compute the probability that a person who tests positive actually has the disease, we can use Bayes' theorem. Let's define the following:

• D: Person has the disease
• P: Person tests positive

We are given:

• P(D) = 0.005 (incident rate)
• P(¬D|P) = 0.04 (false negative rate)
• P(P|¬D) = 0.01 (false positive rate)

Using Bayes' theorem:

P(D|P) = [P(D) * P(P|D)] / [P(D) * P(P|D) + P(¬D) * P(P|¬D)]

Substituting the given values:

P(D|P) = [0.005 * (1 - 0.04)] / [0.005 * (1 - 0.04) + (1 - 0.005) * 0.01]

Simplifying the expression:

P(D|P) = 0.00495 / (0.00495 + 0.995 * 0.01) = 0.00495 / 0.01495 = 0.3311

Therefore, the probability that a person who tests positive actually has the disease is approximately 0.3311, or 33.11%.