Question

Suppose there is a disease that affects 1% of the population. Researchers developed a diagnostic test for this disease. The test has a sensitivity of 95% (meaning it correctly identifies 95% of people with the disease) and a specificity of 90% (meaning it correctly identifies 90% of people without the disease). If a person tests positive for the disease, what is the probability that they actually have the disease, according to Bayes' Theorem?

A) Approximately 9.5%

B) Approximately 19%

C) Approximately 50%

D) Approximately 90%

Answer 1

According to Bayes' Theorem, the probability that a person actually has the disease given that they tested positive can be calculated using the formula: P(Disease | Positive) = (P(Positive | Disease) imes P(Disease)) / P(Positive). In this case, the probability is approximately 9.1%.

Step-by-step explanation:

According to Bayes' Theorem, the probability that a person actually has the disease given that they tested positive can be calculated using the formula:

P(Disease | Positive) = (P(Positive | Disease) imes P(Disease)) / P(Positive)

In this case, the probability of having the disease is 1%, or 0.01. The sensitivity of the test is 95%, or 0.95, and the specificity is 90%, or 0.9. Therefore, the probability of a positive test result given that the person actually has the disease is 0.95, and the probability of a positive test result given that the person does not have the disease is 0.1 (1 - specificity). Finally, the probability of a positive test result is calculated as follows:

P(Positive) = (P(Positive | Disease) imes P(Disease)) + (P(Positive | No Disease) imes P(No Disease)) = (0.95 imes 0.01) + (0.1 imes 0.99) = 0.1045

Substituting these values into the formula, we get:

P(Disease | Positive) = (0.95 imes 0.01) / 0.1045 ≈ 0.091 or 9.1%