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A medical clinic tests blood for certain disease from which approximately one person in a hundred suffers. People come to the clinic in group of 50. The operator of the clinic wonders whether he can increase the efficiency of the testing procedure by conducting pooled tests. In the pooled tests, the operator would pool the 50 blood samples and test them altogether. If the pooled test was negative, he could pronounce the whole group healthy. If not, he could then test each person‘s blood individually. The expected number of tests the operator will have to perform if he pools the blood samples are:

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    You are right, sorry about that. But in that case, the order in which you do the tests matters, as you have to first have 49 tests that come negative and finally the 50th test unneeded. The relevance of this order will probably complicated the solution, and the problem states that the operator will test each one individually. Maybe that means all 50 tests will happen anyway. I don't know.2012-08-04

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If $p=1/100$ is the probability of an individual suffering from this disease, then the probability of none of the $n=50$ people suffering from it is $(1-p)^n$. Thus, the expected number of tests in the pooled approach is $1+(1-(1-p)^n) n=n+(1-n(1-p)^n)$, compared to $n$ in the direct approach, so the pooled approach is worthwhile if $n(1-p)^n\gt1$, or $p\lt1-n^{-1/n}$. For $n=50$, we have $50^{-1/50}\approx0.075$, so the pooled approach works for $p=1/100$.

[Edit in respone to a comment under the question:]

As you rightly pointed out, this misses the special case where the batch test is positive and the first $n-1$ individual tests are negative. The probability for that to happen is $(1-p)^{n-1}p$, so the expected number of tests for the batch approach is actually

$ \begin{align} 1+\left(1-(1-p)^n\right)n-(1-p)^{n-1}p &= n+\left(1-n(1-p)^n-(1-p)^{n-1}p\right) \\ &=n+\left(1-(n(1-p)+p)(1-p)^{n-1}\right)\;. \end{align} $

The condition for the added term to be negative can no longer be solved for $p$ in closed form. With your numbers, the expected number is

$50+\left(1-\left(50\left(1-\frac1{100}\right)+\frac1{100}\right)\left(1-\frac1{100}\right)^{50-1}\right)\approx21\;.$

The substracted term $(1-p)^{n-1}p$ is just

$\frac1{100}\left(1-\frac1{100}\right)^{50-1}\approx0.006\;,$

so it doesn't make an appreciable difference. The ratio between the two subtracted terms is $n(1-p)/p=50\cdot99=4950$.

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    Sorry I posted it all wrong2012-08-08