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I have been trying to find out below problem;

Suppose 10 patients are to be tested for a blood disease and that the test in guaranteed to detect the disease. Furthermore, suppose that the probability that a patient has the disease is 0.02 and that presence of the disease in any patient is independent of all other patients. One option is to test all patients individually and that would require 10 tests. Another option is to combine blood samples from the 10 patients and test the combined sample. If the test is negative, then all patients are declared disease free and only 1 test is required. If, however, the combined sample tests positive, then all 10 patients are tested individually, requiring a total of 11 tests. In the latter option derive the probability distribution of , the number of tests, and determine Expected Value?

Among all the distribution models, it seems that geometric distribution seems best fit. But, when I use geometric distribution pmf and its' EV=1/p , I get unreasonable answers.

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Two possibilities: Case (A) no patients have the disease. The probability for this situation is $p_A=(1-0.02)^n$ and number of tests $N_A=1$. Case (B) at least one patient has the disease. Can you work out $p_B$? Hint: there is no other case.

The expected value is $E[N]=p_AN_A+p_BN_B$

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  • $\begingroup$ Wow! I think I got it. It would be the complement of probability of nobody having the disease, right? Also the question asks for the probability distribution of number of tests too. How do we do that? $\endgroup$ Dec 3, 2017 at 1:22
  • $\begingroup$ Also I noticed that your expected value equation is based on the fact that there is a fixed number of tests applied for case B. But, I think the question indicates that you stop applying tests once you found the person with disease. $\endgroup$ Dec 3, 2017 at 1:43
  • $\begingroup$ Actually, the question states that if the combined blood test is positive then all 10 patients are tested individually. It is possible that two patients are positive, so all should be tested. The distribution would have two possible outcomes, 1 test or 11 tests, the probability for each outcome is $p_A$ and $p_B$. $\endgroup$
    – Dean
    Dec 3, 2017 at 18:27
  • $\begingroup$ I see. What confused me was " In the latter option derive the probability distribution of , the number of tests, and determine Expected Value?" I thought that the number of tests would be a random variable and we would have to find the distribution of that. $\endgroup$ Dec 3, 2017 at 19:06
  • $\begingroup$ You are right. The number of tests is a random variable and you have to find the distribution of that. The distribution is not a "famous" one, rather it is described by $f(n)=p_A$ for $n=1$ and $f(n)=p_B$ for $n=11$ and $f(n)=0$ otherwise. That is a probability mass function (or more simply called a distribution). $\endgroup$
    – Dean
    Dec 3, 2017 at 20:50

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