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Consider a disease whose presence can be identified by carrying out a blood test. Let \(p\) denote the probability thar 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 \(n\) blood samples. 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 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 \(p=.1\) and \(n=3\), what is the expected number of tests using this procedure? What is the expected number when \(n=5\) ? [The article "Random Multiple-Access Communication and Group Testing" (IEEE Trans. Commun., 1984: 769-774) applied these ideas to a communication system in which the dichotomy was active/ idle user rather than diseased/nondiseased.]

Step by step solution

01

Understanding the Problem

We need to find the expected number of tests when using group testing for disease detection. If a combined test is negative, we stop after one test. If positive, we perform additional tests for each individual.

02

Defining Events and Outcomes

Consider the event that a person has the disease (probability \( p \)) and that a person does not have it (probability \( 1-p \)). For \( n \) individuals, group testing is used first, followed by individual testing if necessary.

03

Calculating Probability for Combined Negative Result

The probability that none of the \( n \) individuals have the disease (resulting in a negative combined test) is \((1-p)^n\).

04

Calculating Probability for Combined Positive Result

The probability that at least one individual has the disease (resulting in a positive combined test) is \( 1 - (1-p)^n \). This requires \( n+1 \) tests to identify the diseased individuals.

05

Expected Number of Tests for \( n=3 \)

The expected number of tests \( E(T) \) is the sum of probabilities times their respective number of required tests:\[E(T) = 1 \times (1-p)^3 + (3+1) \times [1-(1-p)^3]\]Substitute \( p = 0.1 \) and \( n = 3 \):\[E(T) = 1 \times (0.9)^3 + 4 \times [1-(0.9)^3] = 1 \times 0.729 + 4 \times (1 - 0.729) = 1.081\]

06

Expected Number of Tests for \( n=5 \)

Using the same logic as before, substitute \( n = 5 \):\[E(T) = 1 \times (0.9)^5 + 6 \times [1-(0.9)^5]\]Calculate this:\[E(T) = 1 \times (0.59049) + 6 \times (1 - 0.59049) = 1 \times 0.59049 + 6 \times 0.40951 = 2.04755\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability in Group Testing

Probability is a foundational concept in group testing and helps us understand the chances of different outcomes during the testing process. In this context, we're dealing with probabilities related to whether each individual has a disease or not.

• Suppose the probability that a randomly selected individual has the disease is represented by the variable \( p \). This is a fixed number that describes how prevalent the disease is in the population to be tested.
• If \( p = 0.1 \), then this means there is a 10% chance that any randomly selected individual has the disease.
• The probability that an individual does not have the disease would be \( 1-p \), complementing the probability that they do have the disease.

To advance from individual to group testing, we are interested in the probability that none of the \( n \) group members have the disease, and the test returns negative. This is given by the equation \((1-p)^n\). Thus, the probability that at least one person in the group has the disease would be \( 1 - (1-p)^n \).
Comprehending these probabilities is key because they determine how many tests potentially need to be done.

Expected Value in Group Testing

Expected value is a crucial measure in probability theory, often used to predict the average outcome of a random process like group testing. It represents the average number of tests required when the group testing method is applied repeatedly under the same conditions.

In our exercise, we need to compute the expected number of tests using the following formula:

• For a group that tests negative, only one test is needed. Hence, the expected value can be calculated as \(1 \times (1-p)^n\).
• On the other hand, if at least one individual tests positive, there will be one group test plus \(n\) individual tests, requiring \(n+1\) tests.
• The expected number of tests for this scenario will be \((n + 1) \times [1-(1-p)^n]\).

Thus, the overall expected number of tests is obtained by summing the products of the probabilities and their corresponding test numbers: \[E(T) = 1 \times (1-p)^n + (n+1) \times [1-(1-p)^n]\]
This equation helps us understand how the expected number of tests changes with different group sizes \( n \) and disease probabilities \( p \), providing insights into the efficiency of group testing strategies.

Disease Detection using Group Testing

In disease detection, group testing is a strategy to efficiently identify individuals with a disease, such as through blood tests. This method can significantly save on resources when testing a large number of individuals.

• The idea is to pool samples from \( n \) individuals and perform a single test on the combined sample. This reduces the number of tests when most results are negative.
• If the pooled test result is negative, it means none of the individuals have the disease, and only one test is needed.
• If positive, each individual in the group needs to be tested separately to identify who carries the disease.

This process was notably used during World War II to identify syphilitic soldiers efficiently. It aligns with modern applications in which minimizing the number of tests is crucial, both economically and logistically, particularly in large populations where the disease prevalence is relatively low.

Independent Events in Group Testing

Independent events are pivotal in understanding the probabilistic model of group testing. Events are independent if the occurrence or non-occurrence of one does not affect the other.

In the context of group testing for disease detection:

• We assume each individual's disease status is independent of others, meaning whether one person tests positive or negative does not influence the probability that another person will test positive or negative.
• This independence simplifies calculations, as the joint probability of \( n \) individuals each not having the disease is \((1-p)^n\), simply multiplying the individual probabilities together.

This assumption underpins the group testing strategy where each test is considered individually, and the group's result is a compound of independent individual outcomes. Understanding independence ensures that the testing results are accurate and that statistical methods applied are valid, thereby effectively managing the testing process.