I had no idea where to even start with this question. Please help.

Two doctors independently examine a person randomly chosen from a certain population to check for the presence or absence of a particular disease. Let CA be the event that Doctor A makes the correct diagnosis, 03 the event that Doctor B makes the correct diagnosis, and D the event that the randomly chosen patient actually has the disease in question. The doctors are equally skilled, so we have P(CAlD) = P(CBlD) 2 p1 and P(C'AlDC) = P(CBIDC) 2 130 Finally, let 6 = P(D) be the prevalence of the disease in the population, that is, the probability the patient has the disease. (a) Prove that the probability both doctors make the correct diagnosis is given by 6(1)? 103) + 193- (b) Show that, in general, P(C'Bch) 75 P(CB). What do you conclude? How would you explain this to a non-statistician? (c) Find all the possible values of 6, p0, and p1 for which we have P(CBch) = P(CB). What special cases do these correspond to