Question

A test has been developed to diagnose certain disease. The following information is available:

• 0.6 % of the population have the disease
• When a person has the disease, the probability that the test gives a (+) signal is 0.96
• When a person does not have the disease, the probability that the test gives a (-) signal is 0.04

a)      If your test result is (+), what is the probability that you actually have the disease?

b)     If your test result is (-), what is the probability that you actually have the disease?

c)      For many medical tests, it is standard procedure to repeat the test when a positive signal is given. If repeated tests are independent, what is the probability that you will test (+) on two successive tests if you have the disease?

d)      If you test (+) on two successive tests, what is the probability that you have the disease?

(Hint: use Bayes theorem)

Answer 1

Let D denote that person has dire are P (D) P(OC) Then 0.006 --- 0.994 Also -0.96 and given P(+1D) P(-10) 0.04 P(-10) 0.04 and P(+10%)=0.96 a) PDI +) P (+1D) PCD) P(+1D) P(D) + P(+18) P (DC) - 0.96% 0.006 0.96 0.006 + 0.96*0.494 0.00576 - 0.006 0.00576 t 0.95424 b) P(D1-) = P(-1D) P(D) PC-ID) P()+PC-1) (D) = 0.006 0.04 x 0.006 0.024 X 0.006 +0.04 x0.994 P(+1D) =0.96 tests that are So, for two successive independent probability of testing (+) & is P(++/D)= 0.98 %0.96 = 0.9216 d) Similar to abone P(++/D") = 0.96*0.96 0.9216 PCD I++) 1 = P(++/D) P(D) P(++D) P(D) +PC++D9P(D)

О.4216 кооо, So, РСъ| ++) 0,42 16х0.00 +0.92) 6х1994 О.00 С.