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Suppose that 5% of all Olympic athletes use performance-enhancing drugs. Because drug use is a serious violation of Olympic rules, the International Olympic Committee has implemented random drug tests. Assume that the test will report positive with probability 0.8 if the athlete uses drugs, and with probability 0.2 if the athlete does not use drugs.

1) Typically, an athlete is asked to give two test samples, A and B, for one drug test. Assume that the results of A and B are run independently, if both of an athlete’s samples are positive, what is the probability that he uses drugs?

So far I have P(positive) = 0.8 for both A test and B test, so the P(A intersecting B) = 0.8 * 0.8 = 0.64 (or 64%). I know I need to do something with the 5% chance that all athletes use drugs, but I'm not sure where do go from here. Is it the probability of the intersection of A and B and use drugs? I'm completing this question for a course and have changed the numbers so it's not exact, but I'm having trouble coming up with the equation I need to solve this.

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You're close.

Let $D$ be the event that an athlete takes drugs, $P_A, P_B$ the events that the athlete tests positive for tests $A$ and $B$, respectively.

We want to find: $P(D|P_A,P_B)$. We know from Bayes rules that: $$P(D|P_A,P_B) = \frac{P(P_A,P_B|D)P(D)}{P(P_A,P_B)}$$

You know $P(D)$ and $P(P_A,P_B|D)$. So know you need to find $P(P_A,P_B)$. Give it a shot and let us know if you need more help.

Edited

Note that your assertion:

P(positive) = 0.8 for both A test and B test

Is not technically correct. The question states the test will report positive with probability 0.8 if the athlete uses drugs. That's $P(P_A|D)$ or $P(P_B|D)$. That's different than $P(P_A)$ or $P(P_B)$.

But your treatment of them being independent is correct, per A and B are run independently.

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  • $\begingroup$ So then if P(Pa|D) = 0.8, is P(Pa,Pb|D) also 0.8 because they have the same probability of occurring, i.e. 80% chance that the athlete will test positive for both A and B while being on drugs? And to find P(Pa,Pb) I will need to find the probability of each event occurring at the same time? $\endgroup$
    – user131935
    Sep 22, 2016 at 2:02
  • $\begingroup$ No. You were correct above. My comment is only to indicate that your statement about why that's the case, isn't technically accurate. So P(Pa,Pb|D)=.64 $\endgroup$
    – ilanman
    Sep 22, 2016 at 2:06

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