Use Bayes' rule to find the indicated probability.
-The incidence of a certain disease on the island of Tukow is 4%.A new test has been developed to diagnose the disease.Using this test,91% of those who have the disease test positive while 4% of those who do not have the disease test positive (false positive).If a person tests positive,what is the probability that he or she actually has the disease?
A)0.438
B)0.91
C)0.487
D)0.856

Question

Quizplus · Accepted Answer

Let A be the event that a person has the disease and B be the event that a person tests positive.

We want to find P(A|B), the probability that a person has the disease given that they tested positive.

Using Bayes' rule:
P(A|B) = P(B|A)P(A)/P(B)

We know:
P(A) = 0.04 (given)
P(B|A) = 0.91 (given)
P(B|not A) = 0.04 (given)

To find P(B), we can use the law of total probability:
P(B) = P(B|A)P(A) + P(B|not A)P(not A)
= 0.91*0.04 + 0.04*0.96
= 0.0764

Now we can substitute these values into Bayes' rule:
P(A|B) = 0.91*0.04/0.0764
= 0.487

Therefore, the probability that a person actually has the disease given that they tested positive is 0.487 or approximately 0.49.

Use Bayes' Rule to Find the Indicated Probability