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

Answer:

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

Step-by-step 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:

$P(D|P)=\frac{P(D)\cdot P(P|D)}{P(D)\cdot P(P|D)+P(\overline{D})\cdot P(P|\overline{D})}\\P(D|P)=\frac{0.006\times 0.92}{0.006\times 0.92+0.994\times 0.05}\\P(D|P)=\frac{5.52\times 10^{-3}}{5.52\times 10^{-3}+49.7\times 10^{-3}}\\P(D|P)=\frac{5.52}{55.22}\\P(D|P)=0.0996=0.0996\times 100=9.96\%$

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