Question

A certain disease has an incidence rate of 0.8%. If the false negative rate is 6% and the false positive rate is 3%, compute the probability that a person who tests positive actually has the disease.Give your answer accurate to at least 3 decimal places

Answer 1

SOLUTION:

Case: Probability

Given:

Incidence rate = 0.8%

False negative = 6%. hence True negative is = (100% - 6%) which is 94%

False positive = 3%, hence true positive is = (100% - 3%) = 97%

FN = False negative

TN= True negative

FP= False positive

TP= True positive

Required: To find the probability that a person who tests positive actually has the disease.

Method:

The chances that someone who test positive actually has the diseases will be given as:

$\begin{gathered} Pr(TP)\text{ = }\frac{n(TP)}{n(TP\text{ + FP)}} \\ Pr(TP)\text{ = }\frac{97}{97+\text{ 3}} \\ Pr(TP)\text{ = 97\%} \\ Pr(TP)\text{ = }0.97 \end{gathered}$

Final answer:

The probability that a person who tests positive actually has the disease is 0.970