Question

A screening test for a rare form of TB has a 7% false positive rate (i.e.
indicates the presence of the disease in people who do not have it). The
test has an 8% false negative rate (i.e. indicates the absence of the disease
in people who do have it). In a population of which 0.6% have the disease,
what is the probability that someone who tests positive has the disease?

Answer 1

ANSWER:

(i) Probability of having disease, P(B1) = 0.006.

Probability of not having disease, P(B2) = 1 - P(B1)

P(B2) = 1 - 0.006

= 0.994.

(ii) P(testing positive | having disease), P(A | B1) = 1 - P(false negative)

= 1 - 0.08

= 0.92

P(testing positive | not having disease), P(A | B2) = P(false positive) = 0.07

P(A) = P(A B1) + P(A B2) = P(A | B1) x P(B1) + P(A | B2) x P(B2)

=( 0.92 x 0.006 )+ (0.07 x 0.994 )

= (0.00552+0.06958)

=0.0751.

(iii) P(has disease | tests positive),

P(B1 | A)

=P(A B1) / P(A)

=(0.92*0.006)/(0.0751)

=0.0735