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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 \(n\) 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. on 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

Define the Problem

We need to calculate the expected number of tests required when pooling blood samples of individuals in a group and retesting individually if the test on the pooled samples is positive. The probability that an individual has the disease is 0.1 (\(p=0.1\)), and we will calculate for two scenarios: Group sizes \(n=3\) and \(n=5\).

02

Determine the Probability of a Negative Test on the Combined Sample

The probability of a negative test on the combined sample is the probability that none of the individuals in the group has the disease. This is given by \((1-p)^n\). For \(n=3\), it becomes \((0.9)^3\) and for \(n=5\), it becomes \((0.9)^5\).

03

Calculate Probability Values for Both Scenarios

For \(n=3\), the probability of a negative test is \((0.9)^3 = 0.729\). For \(n=5\), the probability of a negative test is \((0.9)^5 \approx 0.59049\).

04

Compute the Expected Number of Tests for Group N=3

The expected number of tests, \(E(T)\), for \(n=3\) is calculated as \(E(T) = 1 \cdot P(\text{negative test}) + (1 + n) \cdot P(\text{positive test})\). For \(n=3\), this becomes \(E(T) = 1\cdot0.729 + 4\cdot(1-0.729)\).

05

Simplify the Expected Number Calculation for N=3

For \(n=3\), the expected number is \(E(T) = 0.729 + 4\times0.271 = 0.729 + 1.084 = 1.813\).

06

Compute the Expected Number of Tests for Group N=5

Similarly, calculate \(E(T)\) for \(n=5\): \(E(T) = 1\cdot0.59049 + 6\cdot(1-0.59049)\).

07

Simplify the Expected Number Calculation for N=5

For \(n=5\), \(E(T) = 0.59049 + 6\times0.40951 = 0.59049 + 2.45706 = 3.04755\).

08

Final Step: Conclusion

The expected number of tests when \(n=3\) is about 1.813, and for \(n=5\), it is about 3.048.

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Key Concepts

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

Probability

Probability is a fascinating concept that helps us quantify the chance of an event happening.
For our problem, this represents the probability (\(p\)) that one randomly selected individual has a disease.
In this scenario, we have p=0.1, meaning there is a 10% chance for any person to have the disease.

In group testing for multiple individuals (for example, 3 or 5), the probability helps determine the likelihood of everyone in the group being disease-free.
This is calculated by the formula \((1-p)^n\), where \(n\) is the number of people in the group.

• For 3 people, the probability no one has the disease is \(0.9^3 = 0.729\). This means there's a 72.9% chance the group test will be negative.
• For 5 people, it becomes \(0.9^5 \approx 0.59049\), or a 59.05% chance.

Understanding these probabilities is pivotal when planning group testing strategies.

Expected Value

The expected value is a key concept in statistics used to predict the average outcome of a random event.
It is like finding a 'long-term average' if we could repeat the process indefinitely.

In group testing, the expected number of tests required, \(E(T)\), showcases how many tests we need on average.
Even though we calculate this once, it's based on scenarios where the combined test may either be negative or positive.

• For group size 3, the expected number is first determined by the probability of a negative test (i.e., all clear): \(E(T) = 1 \cdot P(\text{negative test}) + (1 + n) \cdot P(\text{positive test})\)
• In our case, it becomes \(0.729 + 4 \cdot (1-0.729) = 1.813\) tests on average.
• For group size 5, it becomes \(E(T) = 0.59049 + 6 \cdot (1-0.59049)\), meaning about 3.048 tests.

These values efficiently summarize resources needed for group testing.

Disease Testing

Disease testing helps in identifying individuals who carry a specific disease.
Traditional methods involve testing each person separately, but in crowded or resource-limited settings, this can be costly and inefficient.

Group testing optimizes this process by testing pooled samples first.
If nobody has the disease, only one test is required. All members are then considered healthy.
If the test is positive, it indicates that at least one person could be affected. In that case, individual tests follow.

• This method was originally developed for syphilis testing during World War II.
• More modern applications include communication systems and identifying active users efficiently.

Overall, group testing offers a smart, cost-effective solution in scenarios where large numbers are involved, enabling us to utilize resources judiciously.