chanyingsauu7

2021-11-17

A large number N of people are subjected to a blood investigation. This investigation can be organized in two ways.
(1) The blood of each person is investigated separately. In this case N analyses are needed.
(2) The blood of k people are mixed and the mixture is analysed. If the result is negative, then this single analysis is sufficient for k persons. But if it is positive, then the blood of each one must be subsequently investigated separately, and in toto for k people, k+1 analysis are needed. It is assumed that the probability of a positive result (p) is the same for all people and that the results of the analysis are independent in the probabilistic sense.
For what k is the minimum expected number of necessary analysis attained?

### Answer & Explanation

Tamara Donohue

Given:
Let N represents negative, P represents positive
In a sample of k people, the test comes negative only if none of the k people's blood tests positive.
So, $P\left(N\right)={\left(1-p\right)}^{k}$
The total test population is divided into N/k groups of k people each
Expected Number of test, E(X) is:
$E\left(X\right)=\frac{N}{K}\left(P\left(N\right)+kP\left(P\right)\right)$
$E\left(X\right)=\frac{N}{K}{\left(1-p\right)}^{k}+N\left(1-{\left(1-p\right)}^{k}\right)$
$E\left(X\right)=N\left(1+\frac{{\left(1-p\right)}^{k}\left(1-k\right)}{k}\right)$
Now, for the minimum number of tests, differentiate above equation w.r.t. k,
$\frac{\partial }{\partial k}\left(\frac{{\left(1-p\right)}^{k}\left(1-k\right)}{k}\right)=\frac{1-k}{k}{\left(1-p\right)}^{k}\mathrm{ln}\left(1-p\right)-\frac{{\left(1-p\right)}^{k}}{{k}^{2}}$
$=\frac{{\left(1-p\right)}^{k}}{{k}^{2}}\left(k\left(1-k\right)\mathrm{ln}\left(1-p\right)-1\right)$
We will equate this result to zero for minimum condition,
$k\left(k-1\right)\mathrm{ln}\left(1-p\right)+1=0$
Approximating the logarithms by $\mathrm{ln}\left(1-p\right)\approx -p$ for small p, we obtain
${k}^{2}-k-\frac{1}{p}=0,$
Solving this quadratic, we get
${k}_{1}=\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{1}{p}},{k}_{2}=\frac{1}{2}-\sqrt{\frac{1}{4}+\frac{1}{p}}$
For small p, both solutions are close and given by
${k}_{1}\approx {k}_{2}\approx \sqrt{\frac{1}{p}}$

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