Question

Penalty Worksheet – Due 10-19                      Baye’s Rule          Name____________________________

Instructions:You must show your work and all work must be organized and easy to follow. You will have to make an appointment with me to discuss your work. You will not receive credit if you cannot adequately explain your work to me in person.

Suppose that a disease afflicts 2% of the population. A test correctly tests positive for 95% of affected individuals, and correctly tests negative for 90% of individuals unaffected by the disease. Suppose that the test is routinely administered to individuals,                                                          

a.What is the probability a randomly tested person tests positive for the disease?

b.If a randomly tested person tests positive, what is the probability they actually have the disease?

c.For many medical tests, it is standard procedure to repeat the test when the first test comes back positive. If repeated tests are independent, what is the probability that someone withthe disease tests positive twice? If repeated tests are independent, what is the probability that someone withoutthe disease tests positive twice?

d.If a randomly tested person tests positive on two successive tests (see part (c) for explanation), what is the probability they actually have the disease?

Answer 1

Let P denote the event that the test is positive

and F denote the event that the test is negative

D denote the event that the individual is affected by the disease.

Pr(P|D) = 0.95

Pr(P|D') = 0.90

Pr(D) = 0.02, thus Pr(D') = 1 - Pr(D) = 0.98

(a) Probability that a randomly tested person tests positive for the disease ,Pr(P) = Pr(D)*Pr(P|D) + Pr(D')*Pr(P|D') = 0.02*0.95 + 0.98*0.90 = 0.901

(b) Pr(D|P) = Pr(P|D)*Pr(D)/Pr(P) = 0.95*0.02/0.901 = 0.0211

(c) Probability that someone with the disease tests positive twice, Pr(2P|D) = Pr(P|D)*Pr(P|D) = 0.9025

Probability that someone without the disease tests positive twice,Pr(2P|D') = Pr(P|D')*Pr(P|D') = 0.81

(d) Pr(D|tests positive on two successive tests),Pr(D|2P) = Pr(2P|D)*Pr(D)/Pr(2P)

= 0.9025*0.02/(Pr(P)*Pr(P)) = 0.9025*0.2/(0.901*0.901) = 0.222