Suppose 5% of the population have a certain disease.A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease.What is the probability that a randomly selected person has the disease given that this person is testing positive?
A)0.0475
B)0.1425
C)0.95
D)0.9
E)0.3333

Question

Quizplus · Accepted Answer

To solve this, we use Bayes' theorem. Let D be the event that a person has the disease, and T+ be the event that a person tests positive. We want to find P(D|T+), the probability that a person has the disease given that they test positive.P(D|T+) = [P(T+|D) * P(D)] / [P(T+|D) * P(D) + P(T+|D') * P(D')]Where:- P(T+|D) = 0.95 (probability of testing positive given the person has the disease)- P(D) = 0.05 (probability of having the disease)- P(T+|D') = 0.10 (probability of testing positive given the person does not have the disease)- P(D') = 0.95 (probability of not having the disease)Plugging in the numbers:P(D|T+) = (0.95 * 0.05) / (0.95 * 0.05 + 0.10 * 0.95) = 0.0475 / (0.0475 + 0.095) = 0.0475 / 0.1425 = 1/3 ≈ 0.3333Therefore, the probability that a person has the disease given they test positive is approximately 0.3333 or 33.33%.

Suppose 5% of the Population Have a Certain Disease