To determine whether or not they have a certain disease, 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 analyzed 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 disease); 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 disease is 0.07 for all people, independently of each other.

(a) Compute the expected number of tests necessary for each group.

Question

Marin Mosley

Mathematics

8 May, 19:08

To determine whether or not they have a certain disease, 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 analyzed 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 disease); 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 disease is 0.07 for all people, independently of each other.

(a) Compute the expected number of tests necessary for each group.

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Answers (1)

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Cohen Hardin · Answer 1

Step-by-step explanation:

Given P (-ve results) = 0.93 P (+ve results) = 0.07 Let Y = number of test needed for each group

P (X=1) = only one test needed for the group = each 10 people in the group shows - ve result

= 0.93^10

= 0.7374

P (X=11) = At least one person shows the results

= 1 - P (X = none of them shows the result)

= 1 - 0.93^10 = 0.2626

Hence expected Number of test = E (X) = 1 x 0.7374 + 11 x 0.2626

= 3.62 which is approximately 4

Hence the expected number of tests necessary for each group = 4