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Chapter 3: Q103SE (page 138)

Consider a disease whose presence can be identified by carrying out a blood test. Let \({\rm{p}}\) denote the probability that a randomly selected individual has the disease. Suppose \({\rm{n}}\) individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the \({\rm{n}}\) blood samples. A potentially more economical approach, group testing, was introduced during World War \({\rm{II}}\) to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single 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 diseased, the test on the combined sample will yield a positive result, in which case the n individual tests are then carried out. If \({\rm{p = }}{\rm{.1}}\) and \({\rm{n = 3}}\), what is the expected number of tests using this procedure? What is the expected number when \({\rm{n = 5}}\)? (The article 鈥淩andom Multiple-Access Communication and Group Testing鈥� (IEEE Trans. on Commun., \({\rm{1984: 769 - 774}}\)) applied these ideas to a communication system in which the dichotomy was active/ idle user rather than diseased/non-diseased.)

Short Answer

Expert verified

The expected number of results for \({\rm{n = 3}}\) and \({\rm{n = 5}}\) are \({\rm{E(Y) = 1}}{\rm{.813}}\) and \({\rm{E(Y) = 3}}{\rm{.0475}}\) respectively.

Step by step solution

01

Concept Introduction

The described method can be useful in many areas. Let 鈥�

\({\rm{Y = }}\)the number of tests carried out.

The probability that an individual is diseased 鈥�

\({\rm{p = 0}}{\rm{.1}}\)

Which means that the probability that an individual does not have the disease is 鈥�

\({\rm{q = 1 - 0}}{\rm{.1 = 0}}{\rm{.9}}\)

02

Value for \({\rm{E(Y)}}\)

For \({\rm{n = 3}}\), there could be only \({\rm{1}}\) test carried out-when the group test is negative, or \({\rm{4}}\) tests could be caried out - if the group test is positive (first test) and then \({\rm{3}}\) individual tests are carried out, which makes it total of \({\rm{4}}\) tests.

The Expected Value (mean value) of a discrete random variable\({\rm{X}}\)with set of possible values\({\rm{S}}\)and\({\rm{pmf p(x)}}\)is 鈥�

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {\rm{x}} \cdot {\rm{p(x)}}\)

The following holds 鈥�

\(\begin{array}{c}{\rm{E(Y) = 1}} \cdot {\rm{0}}{\rm{.729 + 4}} \cdot {\rm{0}}{\rm{.271 = }}\\{\rm{1}}{\rm{.813}}\end{array}\)

This indicated that the expected number of tests is smaller than \({\rm{3}}\), which would be must without group testing.

03

Value for \({\rm{E(Y)}}\)

For\({\rm{n = 5}}\), there could be only\({\rm{1}}\)test carried out-when the group test is negative, or\({\rm{6}}\)tests could be caried out - if the group test is positive (first test) and then\({\rm{5}}\)individual tests are carried out, which makes it total of\({\rm{6}}\)tests.

In order to find the expected number of tests first compute the following probabilities 鈥�

\(\begin{array}{c}{\rm{P(Y = 1) = P(\{ group test is negative\} ) = P(\{ 5 individuals do not have disease\} )}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{q}} \cdot {\rm{q}} \cdot {\rm{q = 0}}{\rm{.}}{{\rm{9}}^5}{\rm{ = 0}}{\rm{.5905}}\end{array}\)

\({\rm{(1)}}\): the tests are carried out independently.

Since \({\rm{Y}}\) can take only two values, \({\rm{1}}\) and \({\rm{6}}\), the following is true 鈥�

\(\begin{array}{c}{\rm{P(Y = 6) = 1 - P(Y = 1)}}\\{\rm{ = 1 - 0}}{\rm{.5905}}\\{\rm{ = 0}}{\rm{.4095}}\end{array}\)

The Expected Value (mean value) of a discrete random variable\({\rm{X}}\)with set of possible values\({\rm{S}}\)and\({\rm{pmf p(x)}}\)is 鈥�

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {\rm{x}} \cdot {\rm{p(x)}}\)

The following holds 鈥�

\(\begin{array}{c}{\rm{E(Y) = 1}} \cdot {\rm{0}}{\rm{.5095 + 6}} \cdot {\rm{0}}{\rm{.4095 = }}\\{\rm{3}}{\rm{.0475}}\end{array}\)

This indicated that the expected number of tests is smaller than\({\rm{3}}\), which would be must without group testing.

Therefore, the values obtained are \({\rm{1}}{\rm{.813}}\) and \({\rm{3}}{\rm{.0475}}\).

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Most popular questions from this chapter

Starting at a fixed time, each car entering an intersectionis observed to see whether it turns left (L), right (R), orgoes straight ahead (A). The experiment terminates assoon as a car is observed to turn left. Let X = the numberof cars observed. What are possible X values? List five outcomes and their associated X values.

A plan for an executive travellers鈥� club has been developed by an airline on the premise that \({\rm{10\% }}\) of its current customers would qualify for membership.

a. Assuming the validity of this premise, among \({\rm{25}}\) randomly selected current customers, what is the probability that between \({\rm{2}}\) and \({\rm{6}}\) (inclusive) qualify for membership?

b. Again assuming the validity of the premise, what are the expected number of customers who qualify and the standard deviation of the number who qualify in a random sample of \({\rm{100}}\) current customers?

c. Let \({\rm{X}}\) denote the number in a random sample of \({\rm{25}}\) current customers who qualify for membership. Consider rejecting the company鈥檚 premise in favour of the claim that \({\rm{p > 10}}\) if \({\rm{x}} \ge {\rm{7}}\). What is the probability that the company鈥檚 premise is rejected when it is actually valid?

d. Refer to the decision rule introduced in part (c). What is the probability that the company鈥檚 premise is not rejected even though \({\rm{p = }}{\rm{.20}}\) (i.e., \({\rm{20\% }}\) qualify)?

Let X have a Poisson distribution with parameter \({\rm{\mu }}\). Show that E(X) =\({\rm{\mu }}\) directly from the definition of expected value. (Hint: The first term in the sum equals \({\rm{0}}\), and then x can be cancelled. Now factor out \({\rm{\mu }}\) and show that what is left sums to \({\rm{1}}\).)

An educational consulting firm is trying to decide whether high school students who have never before used a hand-held calculator can solve a certain type of problem more easily with a calculator that uses reverse Polish logic or one that does not use this logic. A sample of \({\rm{25}}\) students is selected and allowed to practice on both calculators. Then each student is asked to work one problem on the reverse Polish calculator and a similar problem on the other. Let \({\rm{p = P(S)}}\), where \({\rm{S}}\) indicates that a student worked the problem more quickly using reverse Polish logic than without, and let \({\rm{X = }}\)number of \({{\rm{S}}^{\rm{'}}}{\rm{s}}\).

a. If \({\rm{p = }}{\rm{.5}}\), what is \({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)?

b. If \({\rm{p = }}{\rm{.8}}\), what is \({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)?

c. If the claim that \({\rm{p = }}{\rm{.5}}\) is to be rejected when either \({\rm{x}} \le {\rm{7}}\) or \({\rm{x}} \ge {\rm{18}}\), what is the probability of rejecting the claim when it is actually correct?

d. If the decision to reject the claim \({\rm{p = }}{\rm{.5}}\) is made as in part (c), what is the probability that the claim is not rejected when \({\rm{p = }}{\rm{.6}}\)? When \({\rm{p = }}{\rm{.8}}\)?

e. What decision rule would you choose for rejecting the claim \({\rm{p = }}{\rm{.5}}\) if you wanted the probability in part (c) to be at most \({\rm{.01}}\)?

The number of pumps in use at both a six-pump station and a four-pump station will be determined. Give the possible values for each of the following random variables:

a. T = the total number of pumps in use.

b. X = the difference between the numbers in use atstations 1 and 2.

c. U = the maximum number of pumps in use at either Station.

d. Z = the number of stations having exactly two pumps in use.
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