Question

Suppose we have a pretty reliable test used to diagnose a disease: 98 percent of people with the disease will get a positive test result and 92 percent of healthy people people test negative. Now suppose that the general prevalence of the disease in the population (that is, it's base rate) is 1/150. How likely is it that a random person who gets a positive test result actually has the disease?

Answer 1

To calculate the likelihood of a random person who gets a positive test result actually having the disease, we can use Bayes' theorem. The likelihood is approximately 14.03%.

Step-by-step explanation:

To calculate the likelihood of a random person who gets a positive test result actually having the disease, we can use Bayes' theorem. The theorem states that the probability of an event A happening given that event B has happened is equal to the probability of event B happening given that event A has happened, multiplied by the probability of event A happening, divided by the probability of event B happening.

In this case, event A is the person having the disease, and event B is the person testing positive. The probability of a random person having the disease is 1/150, and the probability of a random person testing positive given that they have the disease is 98%. The probability of a random person testing positive given that they do not have the disease (a false positive) is 8%. Plugging these values into Bayes' theorem:

P(person has the disease | person tests positive) = (P(person tests positive | person has the disease) * P(person has the disease)) / P(person tests positive)

P(person tests positive) = (P(person tests positive | person has the disease) * P(person has the disease)) + (P(person tests positive | person does not have the disease) * P(person does not have the disease))

Using these equations, we can calculate that the likelihood of a random person who gets a positive test result actually having the disease is approximately 14.03%.