Question

13.      A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not present. One percent of the population actually has the disease. Calculate the probability that a person has the disease given that the test indicates the presence of the disease.

Answer 1

Let

Y = positive test result

$D$ = disease is present (and $D'=$not $D$)

Using Baye’s theorem:

$$P(D|Y)=\dfrac{P(Y|D)P(D)}{P(Y|D)P(D)+P(Y|D')P(D')}$$

$$=\dfrac{0.95(0.01)}{0.95(0.01)+0.005(0.99)}\approx0.6574$$