Problem 65 - A First Course in Probability

Question

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? 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?

Step-by-Step Solution

Answer

\(P(\text{positive test}) = 1 - \left(\frac{999}{10^3}\right)^{500}\) \(P(\text{more than one having disease}|\text{positive test}) = \frac{P(\text{positive test}) - P(\text{no one has disease}) - P(\text{exactly one has disease})}{P(\text{positive test})}\) For Jones, \(P(\text{more than one having disease}) = \frac{P(\text{positive test}) - (\frac{999}{10^3})^{500} - \binom{500}{1} (\frac{1}{10^3})^1 (\frac{999}{10^3})^{499}}{P(\text{positive test})}\) For remaining people after Jones's test, \(P(\text{any remaining person having the disease}|i) = 1 - \left(\frac{999}{10^3}\right)^{(500 - (i - 1))}\)

Step 1: Calculate the probability of not having the disease

First, let's find the probability that each individual doesn't have the disease. This is simply the complement of the probability of having the disease: \(P(\text{not having the disease}) = 1 - P(\text{having the disease}) = 1 - \frac{1}{10^3} = \frac{999}{10^3}\).

Step 2: Calculate the probability that no one in the company has the disease

Since there are 500 soldiers and their probabilities of not having the disease are independent, the probability that no one in the company has the disease is \(\left(\frac{999}{10^3}\right)^{500}\).

Step 3: Calculate the probability of a positive test

The probability of a positive test is the complement of the probability that all soldiers don't have the disease. Therefore, the probability of a positive test is: \(P(\text{positive test}) = 1 - P(\text{no one has disease}) = 1 - \left(\frac{999}{10^3}\right)^{500}\) #b) Probability of more than one person having the disease given a positive test#

Step 1: Calculate the probability of only one person having the disease

To find the probability of only one person having the disease, we can use the binomial formula since there are 500 independent trials with each soldier's probability of having the disease being \(\frac{1}{10^3}\). We want to find the probability of exactly one success (having the disease), so the binomial formula is: \(\binom{500}{1} (\frac{1}{10^3})^1 (\frac{999}{10^3})^{499}\)

Step 2: Calculate the probability of more than one person having the disease

To find the probability of more than one person having the disease, we can subtract the probabilities of nobody having the disease \(P(\text{no one has disease}) = \left(\frac{999}{10^3}\right)^{500}\) and only one person having the disease from the probability of the positive test. Thus: \(P(\text{more than one having disease}|\text{positive test}) = \frac{P(\text{positive test}) - P(\text{no one has disease}) - P(\text{exactly one has disease})}{P(\text{positive test})}\)

Step 3: Substitute the probabilities from part (a)

Replace all the probabilities we calculated in part (a) into the formula we derived in step 2. Then, we can calculate the probability that more than one person has the disease given that the blood test is positive. #c) Probability for Jones#

Step 1: Calculate the probability that others have the disease

Since Jones knows he has the disease, the situation is now concerning the other 499 soldiers, who still have a probability of having the disease of \(\frac{1}{10^3}\) each.

Step 2: Calculate the probability that more than one person has the disease for Jones

Using the formula derived in Step 2 of part (b), we find the probability that more than one person has the disease for Jones. #d) Probability for remaining people given Jones tests positive#

Step 1: Calculate the probability for the remaining people up to i-1

Using the binomial formula for the first i-1 people, we calculate the probability that none of them have the disease.

Step 2: Calculate the probability for the remaining people

Given that the first i-1 tests were negative, we can use the same approach as in parts (a) and (b) to calculate the probability of any remaining person having the disease. We substitute the values of the outcomes from the first i-1 people into the equations derived in part (a) and (b) to find the probability as a function of i.

Key Concepts

Binomial Probability Distribution

When we discuss the chances of an event occurring a certain number of times in a given number of trials, we are delving into what is called a binomial probability distribution. A clear-cut real-world example lies in the army company scenario where each soldier has a certain independent probability of having a disease. This reflects the essential binomial scenario, featuring two possible outcomes: 'having the disease' or 'not having the disease', across a fixed number of independent events—500 soldiers, in this case. The formula used to calculate the probability of 'r' successes in 'n' trials is given as \( P(X=r) = C(n,r) * p^r * (1-p)^{n-r} \), where 'C(n,r)' is the number of combinations, 'p' is the probability of success on a single trial, and '(1-p)' is the probability of failure. By plugging in the respective probabilities and the number of soldiers, you calculate the likelihood of any specific number of soldiers having the disease.

Complement Rule

Diving further into probability theory, the complement rule plays a crucial role—it simplifies calculating the probability of an event by instead finding the probability that the event does not occur, and then subtracting this from one. This rule is expressed as \( P(A') = 1 - P(A) \), where 'P(A')' is the probability that 'A' does not happen, and 'P(A)' is the probability that 'A' happens. Back to our example, it's much easier to determine the probability of no soldiers being sick and subtract this from one to find the probability of at least one soldier being sick—otherwise known as a positive test outcome.

Conditional Probability

The concept of conditional probability is pivotal when dealing with scenarios that involve dependent events. Essentially, it's about finding the probability of an event given that another event has occurred. This relationship can be represented as \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), indicating the probability of 'A' occurring provided that 'B' is true. In the soldiers' case, part (b) of the problem asks for the probability of more than one person having the disease, given that the pooled blood test is positive. This is an application of conditional probability, where the prior occurrence of a positive test changes the likelihood of the subsequent event.

Independent Events

Lastly, for understanding probability in various contexts, the notion of independent events can't be overlooked. Two events are considered independent if the occurrence of one does not affect the probability of the other. The multiplication rule for independent events states that \( P(A \cap B) = P(A) * P(B) \) if 'A' and 'B' are independent. For instance, the soldiers undergoing blood tests is a series of independent events, as the result for one soldier’s health status does not influence another’s. When calculating the chances of not a single soldier being infected, the independence of each event allows us to simply raise the individual probability of not having the disease to the power of the total number of soldiers.