Answer to Question #122139 in Statistics and Probability for Sebastian

Question

From the study concluded that 7.3% of a population suffering from osteoporosis (porous bone disease). A new technique was developed to test a large individual sample, which consists of two groups of people. The first group is the one who is known to suffer from osteoporosis, and the second group is the group that is confirmed not to suffer from osteoporosis from tests using the old technique. The result is the probability of someone who is confirmed to have osteoporosis, producing a positive test is 95%. While those not suffering from osteoporosis, the probability resulting in a positive test with the new technique is 2%.

i) What is the probability that a person will produce a positive test for osteoporosis, regardless of whether he has the disease, or not (group 1, and group 2).

ii) What is the probability that someone does not have osteoporosis, given that the test results with this new technique are positive?

Answer 1

Let $\pi =$ the proportion of people suffered from osteoporosis, $T$ denote the event ''positive test''.

Given $\pi=0.073, P(T|\pi)=0.95, P(T|(1-\pi)=0.02.$

(i) What is the probability that a person will produce a positive test for osteoporosis, regardless of whether he has the disease, or not ?

$$P(T)=\pi\cdot P(T|\pi)+(1-\pi)\cdot P(T| (1-\pi))=$$

$$=0.073\cdot 0.95+(1-0.073)\cdot0.02=0.08789$$

ii) What is the probability that someone does not have osteoporosis, given that the test results with this new technique are positive?

From Bayes' Rule

$$P((1-\pi)|T)=\dfrac{(1-\pi)\cdot P(T| (1-\pi))}{\pi\cdot P(T|\pi)+(1-\pi)\cdot P(T| (1-\pi))}=$$

$$=\dfrac{(1-0.073)\cdot0.02}{0.08789}\approx0.210946$$