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In a certain clinic $1$% of patients have a certain virus.

If a test is carried out on a positive patient, there is a $5$% probability that the test will give a false result.

If a test is carried out on a negative patient, there is a $10$% probability that the test will give a false result.

A) What is the probability that a patient has the virus given a positive result?

B) Suppose a second test is carried out on the same patient, what is the probability of two positive results?

I'm stuck with part B

Because the testing is independent (The outcome of the first test doesn't affect the outcome of the second test) does this mean that the probability of getting two positive results is the probability of getting one positive result times $2$?

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  • $\begingroup$ If $p$ is the probability of a positive result then the probability of twice a positive result by two independent tests is not $2p$ (as you suggest) but is $p\times p$. $\endgroup$ – drhab May 16 '18 at 9:35
  • $\begingroup$ @drhab Great, I was on the right track anyway thanks for you answer! $\endgroup$ – Darragh O'Flaherty May 16 '18 at 9:48
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Part B is just the same as Part A, except that instead of using 0.01 as the prior possibility that the patient has the virus, you use the answer to part A.

Bayes theorem is nice this way: you can stack processes togetehr, and the posterior of one becomes the prior of the next.

'Independent' means that the 5% and 10% numbers are the same. If a healthy patient gets a false positive in the first test, this is because of some random effect in the test, not because they are a patient with some predisposition to give a false result. If a patient is really ill then the probability of a positive on the second test is not the same as the forst.

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  • $\begingroup$ So the answer from part a is $$\frac{0.0095 * 0.01}{0.1085}$$. So the answer would be $$\frac{(0.0095) * (\frac{19}{21700})}{0.1085}$$ $\endgroup$ – Darragh O'Flaherty May 16 '18 at 9:40
  • $\begingroup$ Your numbers have some power of ten awry - problem with percentages. First part is 0.95*0.01/.1085=0.0876. Don't forget to change the denominator in the second part. It is no longer .1085 but .95*.0876+0.1*(1-.0876)=.17446. $\endgroup$ – RogerJBarlow May 16 '18 at 9:53
  • $\begingroup$ It is a minutely small number 0.0000007666333963, but it is the correct answer. $\endgroup$ – Darragh O'Flaherty May 16 '18 at 9:55

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