Question

A diagnostic test for a disease has a true positive rate of 98% and a true negative rateof 94%. Suppose the overall disease rate in the population is 1%. That is,•Pr(Positive Test|Patient has Disease) = 0.98,•Pr(Negative Test|Patient does NOT have Disease) = 0.94,•Pr(Patient has Disease) = 0.01.a) What is the probability of a patient without the disease testing positive? b) Find the probability that a patient has the disease given that they tested positive.That is, findPr(Patient has Disease|Positive Test).

Answer 1

Solution:

Given:

Pr(Positive Test|Patient has Disease) = 0.98,

Pr(Negative Test|Patient does NOT have Disease) = 0.94,

Pr(Patient has Disease) = 0.01.

Let TP = Positive Test , TN = Negative Test

D =  Patient has Disease

ND = Patient does NOT have Disease

thus we have:

P( TP | D) = 0.98

P(TN | ND) = 0.94

P(D) =0.01

then P(ND) = 1 - P(D) =1 - 0.01 = 0.99

Part a) What is the probability of a patient without the disease testing positive?

P( TP | ND) = ............?

We have:

P(TN | ND) = 0.94

then

P( TP | ND) = 1 - P(TN | ND)

P( TP | ND) = 1 - 0.94

P( TP | ND) = 0.06

Part b) Find Pr(Patient has Disease|Positive Test)=..........?

P( D | TP )=.............?

Using Bayes rule of probability:

Pr(Patient has Disease|Positive Test) = 0.141618