To determine whether or not they have a certain desease, 100 people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to group the people in groups of 10. The blood samples of the 10 people in each group will be pooled and analized together. If the test is negative. one test will suffice for the 10 people (we are assuming that the pooled test will be positive if and only if at least one person in the pool has the desease); 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 the probability that a person has the desease is 0.05 for all people, independently of each other, and compute the expected number of tests necessary for each group.

Question

Mathematics

Guest

18 July, 12:41

To determine whether or not they have a certain desease, 100 people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to group the people in groups of 10. The blood samples of the 10 people in each group will be pooled and analized together. If the test is negative. one test will suffice for the 10 people (we are assuming that the pooled test will be positive if and only if at least one person in the pool has the desease); 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 the probability that a person has the desease is 0.05 for all people, independently of each other, and compute the expected number of tests necessary for each group.

+4

Answers (1)

Know the Answer?

Jessica Curtis · Answer 1

Answer: Expected number of tests = 5.013

Step-by-step explanation:

Define random variable X that marks the number of tests required for some certain

group. Observe that if the test is negative for all the people (which has probability 0.95), we make one and only one test. If some of the people is tested positive, we make ten additional tests for every person in that group separately. Hence, the expected number of tests will be for if they are all negative (1 test) and the case of at least one person testing positive (11 tests).

That Is,

E (X) = 1 (0.95^10) + 11 (1 - (0.95^10))

E (X) = 0.5987 + 4.414 = 5.013