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 sample. 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 1 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 sick, the test on the combined sample will yield a positive result, in which case the n individual tests are the carried out. If p = .1:
What is the expected number of tests using this procedure when n = 3? Is this procedure better on average than simply testing everyone?

0 tests

Yes, this procedure is better on the average than testing everyone, it makes it less cumbersome.

Step-by-step explanation:

Given the information:

Let P be the probability that a randomly selected individual has the disease = 0.1. N individuals are randomly selected, thereafter, blood samples of each person would be tested after combining all specimens. Should in case one person has the disease then it yields a positive result and test should be set for each person.

Let Y be number tests

For n = 3 there are two possibilities. If no one has the disease then the value is 1 otherwise the value is 4, here P = 0.1

Therefore, for Y = 1

P(Y-1) = P(no one has disease)

= 0.9³

= 0.729

If Y = 4

P(Y-4) = 1-P(y = 1)

= 1 - 0.729 = 0.271

The expected number of tests using this formular gives

E(Y) = 1×0.729 + 4×0.271

E(Y) = 0

answer: 8d

step-by-step explanation:

45+5= 50+5=55+5=60

answer: apothems is your answer

step-by-step explanation: hope u get it right