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Assume that for a randomly selected person:

$P (D) = 0.02$, $P (R\mid D) = 1,$ $P (R\mid D') = 0.05$

So that the inexpensive test only gives false positive, and not false negative, results.

Suppose that this inexpensive test costs 10 dollars.

If a person tests positive then they are also given a more expensive test, costing 100 dollars, which correctly identifies all persons with the disease.

What is the expected cost per person if a population is tested for the disease using the inexpensive test followed, if necessary, by the expensive test?

Beginning with the question, I'm not sure which formula to plug in and use, any help will be appreciated thanks!!

Edit

Let $D = \{\text{person has the condition}\}$

Let $R = \{\text{the test result is positive}\}$

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  • $\begingroup$ What exactly are the events D and R? $\endgroup$ – Math1000 Oct 30 '14 at 18:12
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    $\begingroup$ Sorry edited in! $\endgroup$ – user184692 Oct 30 '14 at 18:18
  • $\begingroup$ I think you should edit the probability too, did you P(R/D') = 0.05 $\endgroup$ – Satish Ramanathan Oct 30 '14 at 18:23
  • $\begingroup$ sorry edited as well! $\endgroup$ – user184692 Oct 30 '14 at 20:21
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$P(D) = .02$

$P(R/D) = 1$

$P(R/D') = .05$

$P(D') = 1-.02 = 0.98$

$P(R) = P(D)P(R/D) + P(D')P(R/D') = .02(1) + .98(.05) = .069$

Cost of inexpensive + follow up test, if tested positive $=10 + 0.069(100)$$

Expected cost $= 10+ 0.069*100 = 16.9 $$

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  • $\begingroup$ Sorry but my answer in the book seems to be $16.90, do you think its wrong? $\endgroup$ – user184692 Oct 30 '14 at 20:15
  • $\begingroup$ They are right, the everyone takes the inexpensive test to find if it is positive, so there is no probability there, but the $ 100 is spent only if it is positive. They are correct it should be 10+ 0.069*100 = 16.9, I will edit it. $\endgroup$ – Satish Ramanathan Oct 30 '14 at 20:48
  • $\begingroup$ Alright i got it, thanks a lot man! $\endgroup$ – user184692 Oct 30 '14 at 20:54

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