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

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.

Case 1: P(X=1) = [P (not infected)]⁵

                       = (0.15 - 0.1)⁵

            P(X=1) = 3.125*10⁻⁷

Case 2: P(X=6) = 1- P(X=1)

                        = 1 - (1 - 0.1)⁵

               P(X=6) = (1 - 3.125*10⁻⁷) = 0.999999

               P(X=6) = 1.0

We can then use the previously determined values to compute the expected number of tests.

E(X) = ∑x.P(X=x)

      = (1).(3.125*10⁻⁷) + 6.(1.0)

 E(X)  =  E(X) = 6.00

Therefore, the expected number of tests, E(X) = 6.00