Question

A certain disease has an incidence rate of 0.6%. If the false negative rate is 8% and the false positive rate is 5%, compute the probability that a person who tests positive actually has the disease.

Answer 1

The probability that a person who tests positive actually has the disease is 9.96 %

Explanation:

Given:

Rate of incidence of disease is 0.6%. So, 0.6% of the persons have actually a disease.

Percent of persons not having the disease will be 100 - 0.6 = 99.4%

Let the event of having disease be 'D'. Therefore,

`P(D)=0.6\% = 0.006\\\\P(\overline{D})=99.4\% =0.994`

False negative rate means that the person having disease shows negative test result.

Let the events 'P' and 'N' represent positive test and negative test results respectively.

As per question,

`P(N|D)=8\%=0.08\\\therefore, P(P|D)=1-0.08=0.92`

Also, false positive test result means the person not having disease showing a positive test result. Therefore,

`P(P|\overline{D})=5\%=0.05`

Now, we are asked to determine the probability of a person who tests positive actually has the disease,
`P(D|P)`, which is given using the Bayes' Theorem:

`P(D|P)=\fracD){P(D)\cdot P(P|D)+P(\overline{D})\cdot P(P|\overline{D})}\\P(D|P)=(0.006* 0.92)/(0.006* 0.92+0.994* 0.05)\\P(D|P)=(5.52* 10^(-3))/(5.52* 10^(-3)+49.7* 10^(-3))\\P(D|P)=(5.52)/(55.22)\\P(D|P)=0.0996=0.0996* 100=9.96\%`

Therefore, the probability that a person who tests positive actually has the disease is 9.96 %