Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly selected individual has the disease. Suppose n individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the n blood samples. A potentially more economical approach, group testing, was introduced during World War II to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the n individual tests are then carried out. If p = .1 and n = 3, what is the expected number of tests using this procedure? What is the expected number when n = 5? [The article “Random Multiple-Access Communication and Group Testing” (IEEE Trans. on Commun., 1984: 769–774) applied these ideas to a communication system in which the dichotomy was active/idle user rather than diseased/nondiseased.]

Solution : Step 1 : A blood test is conducted to find the presence of disease in a group of individuals. For this first, they take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the n individual tests are then carried out. If P= 0.1 and n=3. We have to find the expected number of test to be conducted for this procedure. Step 2 : Here X is the number of tests need to conduct for this procedure. And the probability that a person get affected by the disease =p= 0.1. So the probability that the person is not affected by the disease =1-P= 0.9. It is also given that the experiments are conducted independently. 1) So when the number of individuals is n=3. Consider that the possible values of X are 1 if no one affected or 4 if at least one has the disease. Here The expected number of test will be E(X) = (x=1)*P(x=0)+(x=4)*P(x=4) P(X=1)= probability that no one affected (number of test is one) 3 = (1-0.1) ( experiments are conducted independently) = 0.9^3 =0.729 Similarly P(X=4) = probability that at least one has the disease (number of test is 4) = 1-Probability that no one affected. = 1-P(X=1) = 1-0.729 = 0.271 So expected number of tests in this case E(X) = 0.729+ 4*0.271 = 1.813 2