# Group testing of individuals for disease

The problem asks to find the expected number of tests required to find all the individuals who are infected with a disease. Occurence of the disease has a probability of $0.002$. There were two parts to the question. One in which we divided $1000$ into $10$ of $100$. In that case we can model total number of tests as $10 + 100Y$, where $Y$ is a random variable which denotes if there is any person in a group of $100$ who has the disease. Then the random variable $Y$ will be a binomial random variable with $n = 10$ and $p = 1-(1-0.002)^{100}=0.181$ . In that case the expected number of tests will be $10 + 100*E(Y)= 191$. Now the second part , it says if we do it in two stages ie there are $1000$ people who are divided in $10$ groups of $100$ people and each group of $100$ into $10$ subgroups of $10$ people.

My approach: There are total $100$ subgroups of $10$ people. We can treat it as a binomial with $n = 100$ and $p = 1 - (1-0.002)^{10} = 0.01982$. Therefore the number of people whom we need to test individually is $10*n*p = 19.82$ . Now when we divided the $1000$ people into $10$ groups we need $10$ tests for that, so the total is $29.82$ but the answer given is $31.63$. I am unable to understand, where am I missing.

This test has 3 stages. First you test 10 groups. Each group has 100 people. The probability, that at least one group is positive (infected) is $1-0.998^{100}$. Thus the expected value for the tests of the groups is $10\cdot \left( 1-0.998^{100} \right)$ For each group you have 10 subgroups. The probability, that one of these subgroups is positive tested is $1-0.998^{10}$ You have 100 subgroups. Each subgroup has 10 people. Thus the expected value is $1000\cdot \left( 1-0.998^{10}\right)$