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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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    What did you try?2012-08-04
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    I tried using probability for three cases first case when there are 1 test , 50 test and 51 test. But I am not sure which approach to follow. Please tell me how to proceed2012-08-04
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    You should post what you have so far.2012-08-04
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    Expected probability = $ 1 * (0.99)^50 + 50 * (1 - 0.99)^50 (0.99)^49 + 51* ( 1 - (0.99)^50) (1-(0.99)^49) = 212012-08-04
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    I meant show your work in the original post. You can click edit and use as much space as you need. You're not limited to 600 characters like you are here in the comments.2012-08-04
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    If I understand correctly, there are two cases: 1) The batch test goes well and is the only one needed, or 2) it fails and *an additional* 50 tests must be conducted.2012-08-04
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    There can be one more case, batch test fails, you conduct 49 more test and they come good then you need not conduct test for 50th person as you already know that his test fail2012-08-04
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    I don't believe so. All you know from a pooled test failing is that at least one person is unhealthy. You don't know how many are unhealthy or which ones are. Consider this: I pick any combination of numbers from 1 to 50. You can ask if I've picked at least one person, and I'll tell you Yes or No. My telling you "Yes" doesn't inform you anything about the combination of people I chose, so then you'll have to ask me one by one for each of the 50 people if they are in my selection, and you'll have to exhaust all 50 of them to be certain you haven't missed any.2012-08-04
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    Please see, If I test 49 people and they all come good then 50 has to be bad from piegenhole. You do not need to check . Then 50 test2012-08-04
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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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    What about expected number of tests?2012-08-04
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    Please see there was no such option in the exam.2012-08-08
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    Sorry I posted it all wrong2012-08-08