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I was having some trouble doing some bayesian probability problems and was wondering if I could get any help. I think I was able to get the first two but am confused on the last. If someone could please check my work to make sure I am correct and help me on the last question that would be great.

Here is the problem setup:

Let’s say that we work for the International Olympic Committee (IOC) as part of their Fight Against Doping (https://www.olympic.org/fight-against-doping). We have a drug test for a banned performance-enhancing drug (PED) that is 99.3% accurate at identifying an athlete that has the PED in their system. However, it is only 73% accurate at identifying the absence of PED in the athlete’s system. From a scientific study we also have a strong reason to believe that only 3% of Olympic athletes use this particular PED.

Here is the first question.

1. An athlete tests positive for the PED. Given that the test had a positive result, what is the probability the tested individual uses the PED?

My answer:

P = Positive N = Negative

##p(P|PED) = 0.993##

##p(N|No PED) = 0.73##

##p(PED) = 0.03##

##p(P) = p(P|PED)*p(PED) + p(P|No PED)*p(No PED) = 0.993 * 0.03 + (1 - 0.73) * 0.97 = 0.292##

##p(PED|P) = 0.993 x 0.03 / 0.292 = 0.102##- answer

Here is the second question.

2. As an employee of the IOC, we don’t want to needlessly ban an athlete from the chance to compete in the Olympics. As a result, we decide to institute a protocol that if an athlete tests positive for the use of the PED we will administer a second test. The second test is less accurate at identifying an athlete that has the PED in their system, at only 81%, but is more accurate at identifying the absence of PED in the athlete’s system, with a probability of 90%. If the athlete tests positive for the PED in both the first and second test, what is the probability that the accused individual uses the banned PED? (You may assume the outcome of the second drug test is conditionally independent of the outcome of the first drug test).

My answer:

##P_2## = Positive 2nd Test

##N_2## = Negative 2nd Test

##p(P_2|PED)## = 0.81

##p(N_2|No PED)## = 0.90

##p(PED|P_2) = p(P_2|PED)*p(PED)/p(P_2)##

##p(P_2) = p(P_2|PED)*p(PED) + p(P_2|No PED)*p(No PED) = 0.81*0.03 + (1-0.90)*(1-0.03) = 0.124##

##p(PED|P_2) = p(PED|P_2) * p(PED)/p(P_2) = 0.81*0.03/0.124 = 0.196##

Thus the probability of using PED given that you tested positive for both tests is:

##p(Positive for both) = p(P|PED)*p(P_2|PED) = 0.102*0.196 = 0.0199 \approx 0.20## - answer

Here is question 3 - this is the one I am not sure on.

3. Our information that only 3% of Olympic athletes use the PED came from a

study of 300 athletes. This year we tested 500 athletes and confirmed that 11 of

them used the banned substance. In both cases, only a sample of all athletes to

complete in the Olympics were tested for the PED. As a result there is some

uncertainty, so we decide we would like to express what we have learned as a

probability distribution. In two years when we being testing athletes again, what is the fully specified distribution that we will we use for the percentage of Olympic athletes that use the PED?

I don't get what it means by what the probability distribution would be. Would it be some derivation of a Bayesian posterior distribution with 0.03 as the prior, though I'm not sure what the likelihood or marginal would be.

I know this is the Bayes rule formula:

##p(\theta|x) = p(x|\theta)*p(\theta)/p(x)##

But how would I change this to suit the question?

Apologies for the long post, please let me know if you need any more information or clarification.

Thank you for reading