1.8% of a population is infected with a certain disease.There is a test for the disease; however,the test is not completely accurate.94.2% of those who have the disease will test positive.However 4.8% of those who do not have the disease will also test positive (false positives).What is the probability that a person who tests positive actually has the disease?
A)0.360
B)0.265
C)0.942
D)0.018
E)0.735

Question

Quizplus · Accepted Answer

We can use Bayes' theorem to solve this problem.
Let A be the event that a person has the disease, and B be the event that the person tests positive.
We want to find P(A|B), the probability that a person actually has the disease given that they test positive.
From Bayes' theorem, we have:

P(A|B) = P(B|A)P(A) / [ P(B|A)P(A) + P(B|not A)P(not A) ]

P(B|A) = 0.942 (given)
P(A) = 0.018 (given that 1.8% of the population has the disease)
P(B|not A) = 0.048 (given)
P(not A) = 0.982 (complement of A)

Substituting these values, we get:

P(A|B) = 0.942 x 0.018 / [ 0.942 x 0.018 + 0.048 x 0.982 ]
P(A|B) = 0.265

Therefore, the probability that a person who tests positive actually has the disease is 0.265 or 26.5%. Option B is correct.

18% of a Population Is Infected with a Certain Disease