In the past, an athlete in the Olympic Games who used banned substances had a 99

Question

In the past, an athlete in the Olympic Games who used banned substances had a 99% chance of failing the drug test. For those athletes who do not use banned substances, there is a 95% chance of correctly identifying that they did not use banned substances. Let D be the event that the athlete is a using banned substances, and F is the event that the athlete fails the drug test. Suppose that the International Olympic Committee (IOC) believes that 5% of athletes use banned substances. For a randomly selected athlete, what is the probability that they failed the drug test and used banned substances? Consider, again, the belief of that IOC that 5% of athletes use banned substances. What is the probability that a randomly selected athlete fails the drug test? Consider, again, the belief of that IOC that 5% of athletes use banned substances. For a randomly selected athlete that failed the drug test, what is the probability that this athlete used banned substances? Are the events that an athlete takes banned substances and that an athlete fails the drug test independent? Why or why not? Suppose among a group of 20 Olympic athletes, 6 have never participated in the Olympics before. If we randomly select 4 athletes from this group of 20, what is the probability that at least 1 has never participated in the Olympics before?

Explanation / Answer

Simplest way to solve these types of conditional problems is to use 2 way contingency tables. Without loss of generality, assume there are 10,000 athletes. Using data given, since IOC believes 5% have taken drugs, D has a total of 500 athletes. Of which, 99% fail the test or 495.

a) P(F|D) = 495/970 = 0.51

b) P(Athlete fails test) = 970/10000 = .097

c)This probability is just detection rate or .99. Alternately, req prob = 495/500 = .99

d) No. The 2 events are not independent since there are false positives and false negatives. In other words, test fails to detect 1% of athlete who actually take drugs and flag 5% of those who are clean wrongly.

e) P(Atleast one never participated before) = 1- P(at most one participated before)

                                                              = 1 - P(0) - P(1)

Pick 4 from the 14 who have been to Olympics.

P(0) = 14C4/20C4. = .207

P(1). Pick 1 from the group of 6 that never have and remaining 3 from the group who have

P(1) = 6*14C3/20C4 = .451

Therefore, required probability = 1 - (.207 + .451) = .342

Positive Test Yes No Total Yes 495 475 970 No 5 9025 9030 Total 500 9500 10,000

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