A sports association decides to implement a drug screening procedure to test its
athletes for illegal performance enhancing drugs. A person who does not take the drugs will
test positive with probability 0.02 and a person who does take the drugs will test negative
with probability 0.04. Suppose that 3% of the athletes tested take performance enhancing
drugs. What is the probability that…
a. an athlete testing positive takes the drugs?
b. an athlete testing positive does not take the drugs?
c. an athlete testing negative takes the drugs?
d. an athlete testing negative does not take the drugs?
I was given a hint that I can use Bayes Theorem, but I have no idea how to do this and it looks complicated.. Can anyone help me to solve this..
Best Answer
Refer to the probability tree diagram (D-take drug, D'-do not take drug, P-positive, N-negative):
a) $P(D|P)=\frac{P(D)P(P|D)}{P(D)P(P|D)+P(D')P(P|D')}=\frac{0.0288}{0.0288+0.0194}=0.5975.$
b) $P(D'|P)=\frac{P(D')P(P|D')}{P(D')P(P|D')+P(D)P(P|D)}=\frac{0.0194}{0.0194+0.0288}=0.4025.$
c) $P(D|N)=\frac{P(D)P(N|D)}{P(D)P(N|D)+P(D')P(N|D')}=\frac{0.0012}{0.0012+0.9506}=0.0013.$
d) $P(D'|N)=\frac{P(D')P(N|D')}{P(D')P(N|D')+P(D)P(N|D)}=\frac{0.9506}{0.9506+0.0012}=0.9987.$