Q.4.32 - A First Course in Probability

Question

To determine whether they have a certain disease, $100$ people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to place the people into groups of $10$. The blood samples of the $10$ people in each group will be pooled and analyzed together. If the test is negative, one test will suffice for the $10$ people, whereas if the test is positive, each of the $10$ people will also be individually tested and, in all, $11$ tests will be made on this group. Assume that the probability that a person has the disease is  $.1$ for all people, independently of one another, and compute the expected number of tests necessary for each group. (Note that we are assuming that the pooled test will be positive if at least one person in the pool has the disease.)

Step-by-Step Solution

Answer

The expected number of tests necessary for each group is $7.51$.

Step 1: Given information

$100$ people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to place the people into groups of $10$. The blood samples of the $10$ people in each group will be pooled and analyzed together. If the test is negative, one test will suffice for the $10$ people, whereas if the test is positive, each of the $10$ people will also be individually tested and, in all, $11$ tests will be made on this group. Assume that the probability that a person has the disease is $.1$ for all people, independently of one another

We need to find that the expected number of tests necessary for each group. 

Step 2: Solution

$X=$number of tests conducted.

the probability mass function $X$ is defined as below:

                           $Negative$              $Positive$

$\begin{array}{|ccc|}\hline x:& 1& 11\\ P(X=x):& (1-0.10{)}^{10}=0.{90}^{10}& 1-(0.90{)}^{10}\\ \hline\end{array}$

The wished numeral of tests required for each group is,

$E\left(X\right)=\sum _x xP\left(x\right)$

$=1\times (0.90{)}^{10}+11\times \left[1-(0.90{)}^{10}\right]$

$=0.3487+11\times 0.6513$

$=0.3487+7.1643$

$=7.513$

Step 3: Final answer

The expected number of tests necessary for each group is, $7.51$