Blood is taken from each of n individuals to be tested for a certain disease. Rather than test each sample separately, a pooled method is used in an attempt to reduce the number of tests needed. Part of each blood sample is taken, and these parts are combined to form a pooled sample. The pooled sample is then tested. If the result is negative, then none of the n individuals has the disease, and no further tests are needed. If the pooled sample tests positive, then each individual is tested to see which of them have the disease.

a. Let X represent the number of tests that are carried out. What are the possible values of X?

b. Assume that n = 4 individuals are to be tested, and the probability that each has the disease, independent of the others, is p = 0.1. Find μX.

c. Repeat part (b) with n = 6 and p = 0.2.

d. Express μX as a function of n and p.

e. The pooled method is more economical than performing individual tests if μx<n. Suppose n = 10. For what values of p is the pooled method more economical than performing n individual tests?

Step 1 of 5:

Blood is taken from each of n individuals to be tested for a certain disease. They applied pooled test for this part of each blood sample is taken, and theses parts are combined to form a pooled sample. This pooled sample is then tested. If the result is positive individual test will conduct for each sample. Let X denote the number tests.

We have to find

What are the possible values of XIf n=4 individuals are to be tested, the probability that each has the disease, independent of others, is P= 0.1. What is E(X) Repeat part (b) with n = 6 and P= 0.2 Express E(x) as a function of n and P.The pooled method is more economical than performing individual tests if E(X)< n. Suppose n = 10. For what values of P the pooled method are economical than individual tests.Step 2 of 5:

Consider that if the pooled test is negative there is only one test has to be performed, so X= 1. If pooled test is positive then n additional test has to be carried out , so X= n+1.Therefore the possible values of X are 1 and n+1.

(b) Here n= 4 individuals are to be tested, with probability that each has disease P=0.1.

So there is two chances for the values of X . it can be X=1 when the pooled test is negative. And X= 5 if the pooled test is positive.

Case 1

If X= 1 , since the probability that the selected individual has disease is 0.1.

The probability that none of the 4 individuals has the disease = (1- 0.1

= 0.6561

Case 2

If X=5, that means at least one of 4 has disease.

The probability that at least one of 4 individuals has a disease= 1- no one has disease.

The probability that at least one of 4 individuals has a disease = 1-0.6561

=0.3439 .

So

E(x) = 1 0.6561 + 5 0.3439

= 2.3756.

Therefore the value of E(x) = 2.3756.

Step 3 of 5:

(c) If n= 6, and P=0.2

So there is two chances for the values of X . it can be X=1 when the pooled test is negative. And X= 7 if the pooled test is positive.

Case 1

If X= 1 , since the probability that the selected individual has disease is 0.2.

The probability that none of the 4 individuals has the disease = (1- 0.2

= 0.2621

Case 2

If X=7, that means at least one of 6 has disease.

The probability that at least one of 6 individuals has a disease= 1- no one has disease.

The probability that at least one of 6 individuals has a disease = 1-0.2621

=0.7378 .

So

E(x) = 1 0.2621 + 70.7378

= 5.4267

Therefore the value of E(x) = 5.4267.