Question

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. [The article "Random Multiple-Access Communication and Group Testing"† applied these ideas to a communication system in which the dichotomy was active/idle user rather than diseased/nondiseased.] If p = 0.15 and n = 5, what is the expected number of tests using this procedure? (Round your answer to three decimal places.)

Answer 1

The expected number of tests using the group testing procedure is 1.75.

Step-by-step explanation:

To find the expected number of tests using the group testing procedure, we need to consider the different possible outcomes. Let's break it down:

If no one has the disease (probability = 1 - p), then only one test is required.

If at least one individual has the disease (probability = p), the test on the combined sample will yield a positive result, and then the n individual tests will be carried out.

Therefore, the expected number of tests is:

Expected number of tests = (probability of no disease) * (number of tests in this case) + (probability of disease) * (number of tests in this case)

For the first case, the number of tests is 1.

For the second case, the number of tests is n + 1, because one additional test is required after the positive result from the combined sample.

Substituting the values, Expected number of tests = (1 - p) * 1 + p * (n + 1)

Given p = 0.15 and n = 5, substituting the values we get:

Expected number of tests = (1 - 0.15) * 1 + 0.15 * (5 + 1) = 0.85 * 1 + 0.15 * 6 = 0.85 + 0.9 = 1.75

Therefore, the expected number of tests using this procedure is 1.75.