Question

A scientist designed a medical test for a certain disease. Among 100 patients who have the disease, the test will show the presence of the disease in 96 cases out of 100, and will fail to show the presence of the disease in the remaining 4 cases out of 100. Among those who do not have the disease, the test will erroneously show the presence of the disease in 3 cases out of 100, and will show there is no disease in the remaining 97 cases out of 100.

What is the probability that a patient who tested positive on this test actually has the disease, if it is estimated that 20% of the population has the disease?

Answer 1

Solution:

Let TP = Test Positive and TN = Test Negative

D = Have Disease and ND = No Disease

Thus we have:
P(Test Positive given Disease) = 96/100 = 0.96

That is:

P(TP | D) =0.96

then

P(TN |D) =1 - 0.96 = 0.04

We have :

P(Test Positive given No Disease) = 3/100= 0.03

That is:
P( TP | ND) = 0.03

then

P(TN | ND) =1 -0.03

P(TN | ND) = 0.97

P( Disease) = 0.20

P(D) =0.20

then

P(ND) =1-0.20

P(ND)= 0.80

We have to find:

P( Disease given Test Positive) =........?

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

Using Bayes rule of probability:

$\fn_jvn P( D | TP ) = \frac{P(TP |D) \times P(D)}{P(TP |D) \times P( D)+P(TP |ND) \times P(ND)}$.

$\fn_jvn P( D | TP ) = \frac{0.96 \times 0.20 }{0.96 \times 0.20+0.03 \times 0.80}$

$\fn_jvn P( D | TP ) = \frac{0.192 }{0.192 +0.024 }$

$\fn_jvn P( D | TP ) = \frac{0.192 }{0.216 }$

$\fn_jvn P( D | TP ) =0.888889$

$\fn_jvn {\color{DarkOrange} P( D | TP ) =0.8889}$