Compute the expected number of tests necessary for each grou

Compute the expected number of tests necessary for each group.To determine whether or not they have a certain disease, 200 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 0.04 for all people, independently of each other.

Solution

Compute the expected number of tests necessary for each group.To determine whether or not they have a certain disease, 200 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 0.04 for all people, independently of each other.

Let X be the number of people in the group of 10 who have the disease. X has the binomial distribution with n = 10 trials and success probability p = 0.04


X ~ Binomial( n , p )

n=10, p=0.04

Expectation = np = 0.4

Variance = np(1 - p) = 0.384

Standard deviation = 0.6196773

The test will be positive if X 1

P( X 1) = 1 - P(X = 0) = 1 – 0.6648 = 0.3352


Let Y be the number of groups that will test positive. Y is a binomial random variable with n = 10 trials and successes probability 0.3352

the expectation of Y is: 3.352

Let T be the number of tests administered

There are 20 groups( 200 people to be tested)

E(T) = 20 * E(Y) + 10
E(T) = 20*3.352+10
E(T) = 77.04

We expect 77.04 tests to be given to this group off 200 people.