There are k people taking a blood test at the same time. The blood tests are mixed and analyzed. If the test is positive, all k people will have to do a blood test again but analyzed separately. The probability of a positive result is p. If the number of blood tests equals x, what is the excepted value of x when there are m groups of k people?
In a group of k people, we have at least k blood tests. And if a minimum of one blood test is positive, there are k more blood tests to be done. The probability of having to do another blood test is 1 - $(1-p)^k$. So EX = m(k + (1-$(1-p)^k$)k) = mk(2-$(1-p)^k$). However, the answer is - mk(1 + ${1\over k}$ - $(1-p)^k$). Can anyone tell me why this is right?
Looking at some examples with some actual values for the parameters might help your understanding.
Example Say there are $m = 3$ groups, each containing $k = 2$ people. Suppose any test on an individual's blood has a probability $p = 0.01$ of being positive and is independent of any other individual test.
Then we see that the chance that the pooled test is positive is simply $c = 1 - (1-p)^2 = 0.0199$, because the probability of the complementary event--that both are negative--is $(1-p)^2$.
Furthermore, we must perform at least $m = 3$ pooled tests, one for each group; and with a probability $c = 1 - (1-p)^k = 0.0199$, we must perform an additional $k = 2$ individual tests per positive group. Thus the expected number of total tests is $$\operatorname{E}[X] = 3 + 2(0.0199) = 3.0398.$$
Now can you redo the calculation with general $m, k, p$ to arrive at the correct formula?