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To determine whether they have a certain disease, 100 people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to place the people into groups of \(10 .\) The blood samples of the 10 people in each group will be pooled and analyzed together. If the test is negative, one test will suffice for the 10 people, whereas if the test is positive, each of the 10 people will also be individually tested and, in all, 11 tests will be made on this group. Assume that the probability that a person has the disease is .1 for all people, independently of one another, and compute the expected number of tests necessary for each group. (Note that we are assuming that the pooled test will be positive if at least one person in the pool has the disease.)

Short Answer

Expert verified
The expected number of tests necessary for each group of 10 people is approximately 10.941 tests. This is obtained by considering the different probabilities of people having the disease in a group and the number of tests required in each case. The probabilities are calculated using the binomial formula, and the expected tests for each group are found by adding the products of the probabilities and the number of tests for each outcome.

Step by step solution

01

Calculate the probability of each outcome in a group

For a group of 10 people, the probability of no one having the disease is \((1 - 0.1)^{10}\), and the probability of exactly one person having the disease is given by the binomial formula: \(\binom{10}{1} (0.1)^1 (1 - 0.1)^9\). We can generalize this to find the probability of exactly k people having the disease: \(\binom{10}{k} (0.1)^k (1 - 0.1)^{10 - k}\).
02

Find the expected number of tests for each outcome

If no one has the disease or only one person has the disease, a negative pooled test will be followed by 0 individual tests. For two or more people with the disease, a positive pooled test will be followed by 10 individual tests, totaling 11 tests required for that group.
03

Calculate the expected tests for each group

Now, we can find the expected number of tests for a group by adding the products of the probabilities and the number of tests for each outcome. We have two cases: 1. The case where no one or only one person has the disease: \(E_1 = (1 - 0.1)^{10} \times 1 + \binom{10}{1} (0.1)^1 (1 - 0.1)^9 \times 1\). 2. The case where two or more people have the disease (we sum from k=2 to k=10): \(E_2 = \sum_{k=2}^{10} [\binom{10}{k} (0.1)^k (1 - 0.1)^{10 - k} \times 11]\). Now we add the expected number of tests for the two cases to get the final expected number of tests for each group: \(E = E_1 + E_2\).
04

Compute the values

Calculate values for both cases: \(E_1 = (1 - 0.1)^{10} \times 1 + \binom{10}{1} (0.1)^1 (1 - 0.1)^9 \times 1 \approx 0.736 + 1.468 \approx 2.204\) \(E_2 = \sum_{k=2}^{10} [\binom{10}{k} (0.1)^k (1 - 0.1)^{10 - k} \times 11] \approx 8.737\) Putting these values into the final equation: \(E = E_1 + E_2 \approx 2.204 + 8.737 \approx 10.941\) Thus, the expected number of tests necessary for each group is approximately 10.941 tests.

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Key Concepts

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

Probability Theory
At the heart of predicting outcomes in scenarios like disease testing is probability theory. This branch of mathematics deals with the likelihood of different outcomes occurring. To put it simply, it gives us a way to quantify uncertainty. For example, if we flip a fair coin, the probability of getting heads is 0.5 or 50%, and likewise for tails. In the case of disease testing, probability helps us calculate the likelihood of any individual having the disease and consequently, the expected outcomes of pooled blood tests.

Understanding probability is critical to interpret results and make informed decisions based on them. In our exercise, the individual probability of having the disease is given as 0.1 or 10%. These individual probabilities, when independent, can be combined to figure out the probability of all possible outcomes within a group of people.
Binomial Distribution
A binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states and is especially useful for modeling scenarios where there are a fixed number of identical trials. Diseases testing scenarios, where individuals either have or do not have a disease, map perfectly onto this model. A key feature of the binomial distribution is the use of the binomial coefficient, denoted as \( \binom{n}{k} \), which calculates the number of ways to choose k successes from n trials.

In our exercise, every group of 10 can be thought of as 10 trials, where each trial has a success probability of 0.1 (disease presence). By using the binomial formula to calculate probabilities for various numbers of people having the disease within a group, we can comprehensively understand the distribution of outcomes.
Expected Value Calculation
The expected value is a fundamental concept in probability that provides a measure of the center or 'average' of a probability distribution. Think of it as the long-term average or mean that you would expect if you repeated a random trial many times. The formula for the expected value involves multiplying each possible outcome by its corresponding probability and summing all these products together.

In our exercise, two scenarios are considered—either no one or one person has the disease, or two or more individuals are affected. Calculating the expected value provides us with an average number of tests needed per group, striking a balance between best and worst case scenarios. This allows us to optimize resources by providing an insight into how many tests are needed on average per group.
Disease Testing Optimization
Optimization in disease testing is about finding the most efficient way to use resources to determine disease presence in a population. The goal is to minimize the number of tests without sacrificing accuracy. Pooled testing, as described in our exercise, is one method to optimize testing resources. When individual tests are costly or limited, testing groups of samples can reduce the total number of tests needed, provided the prevalence of the disease (the probability that any individual has the disease) is low.

By pooling together samples, one test can apply to multiple individuals unless a positive result is found, which then requires individual testing. The expected value calculations enable the determination of whether pooling is a cost-effective strategy. Based on the probability of disease presence and the expected value of tests, health administrators can make data-informed decisions on testing strategies.

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Most popular questions from this chapter

Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all \(10 !\) possible rankings are equally likely. Let \(X\) denote the highest ranking achieved by a woman. (For instance, \(X=1\) if the top-ranked person is female.) Find \(\mathrm{P}\\{\mathrm{X}=\mathrm{i}\\}, i=1,2,3, \ldots, 8,9,10\).

In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player, and denote by \(X\) the amount of money he wins in a single game of Two-Finger Morra. (a) If each player acts independently of the other, and if each player makes his choice of the number of fingers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of \(X\) and what are their associated probabilities? (b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of fingers that he guesses his opponent will hold up, and if each player is equally likely to hold up 1 or 2 fingers, what are the possible values of \(X\) and their associated probabilities?

Each of 500 soldiers in an army company independently has a certain disease with probability \(1 / 10^{3} .\) This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested. (a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result. (b) What is the probability, under this circumstance, that more than one person has the disease? Now, suppose one of the 500 people is Jones, who knows that he has the disease. (c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The first \(i-1\) of these tests were negative, and the \(i\) th one-which was on Jones-was positive. (d) Given the preceding scenario, what is the probability, as a function of \(i,\) that any of the remaining people have the disease?

Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter \(\lambda=3\) (a) Find the probability that 3 or more accidents occur today. (b) Repeat part (a) under the assumption that at least 1 accident occurs today.

How many people are needed so that the probability that at least one of them has the same birthday as you is greater than \(\frac{1}{2} ?\)

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