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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? –  ladaghini Aug 4 '12 at 8:58
    
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 proceed –  Arpit Bajpai Aug 4 '12 at 9:11
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You should post what you have so far. –  ladaghini Aug 4 '12 at 9:28
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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. –  ladaghini Aug 4 '12 at 9:37
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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 test –  Arpit Bajpai Aug 4 '12 at 10:32

1 Answer 1

up vote 2 down vote accepted

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? –  Arpit Bajpai Aug 4 '12 at 10:33
    
Please see there was no such option in the exam. –  Arpit Bajpai Aug 8 '12 at 7:48
    
Sorry I posted it all wrong –  Arpit Bajpai Aug 8 '12 at 7:49

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