/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Composite Water Samples The Fair... [FREE SOLUTION] | 91Ó°ÊÓ

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Composite Water Samples The Fairfield County Department of Public Health tests water for the presence of \(E\). coli (Escherichia coli) bacteria. To reduce laboratory costs, water samples from 10 public swimming areas are combined for one test, and further testing is done only if the combined sample tests positive. Based on past results, there is a \(0.005\) probability of finding \(E\). coli bacteria in a public swimming area. Find the probability that a combined sample from 10 public swimming areas will reveal the presence of \(E\). coli bacteria. Is that probability low enough so that further testing of the individual samples is rarely necessary?

Short Answer

Expert verified
The probability is around 0.049 or 4.9%. This suggests further testing of individual samples is not very rare.

Step by step solution

01

Define the Variables

Let us define the probability of finding Escherichia coli bacteria in a single public swimming area as \( p = 0.005 \). The probability of not finding it is \( 1 - p = 0.995 \).
02

Combination Logic

Since we are combining samples from 10 different swimming areas, we need to find the combined probability that all 10 samples don't have E. coli bacteria.
03

Calculate Complement Probability

The probability that none of the 10 samples contain E. coli bacteria is computed using the complement rule. This is given by \( (1 - p)^{10} \).
04

Solve the Complement Probability

Now we perform the calculation: \[ (0.995)^{10} \approx 0.951 \]
05

Calculate the Desired Probability

The probability that at least one sample contains E. coli bacteria is the complement of the previous result. Thus, we have: \[ 1 - (0.995)^{10} \approx 1 - 0.951 = 0.049 \]
06

Interpret the Result

The probability of a combined sample revealing the presence of E. coli bacteria is approximately \(0.049\) or \(4.9\%\). Since this is just under 5%, further testing of the individual samples is not very rare.

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Key Concepts

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

Understanding the Complement Rule
The complement rule in probability helps us find the probability of an event not happening by subtracting the probability of the event happening from 1. If event A has a probability of occurring, denoted as P(A), the complement (the event not occurring) is given by 1 - P(A). In our exercise, the probability of not finding E. coli in one swimming area is 0.995. So, the complement rule is used to find the likelihood of no E. coli in all 10 samples combined. When calculating, we start with the probability of not finding E. coli in one sample and raise it to the power of the number of samples: \(0.995^{10}\). This gives us the overall probability that no sample contains E. coli.
Combining Probabilities
To find the probability of multiple independent events all occurring, we multiply their individual probabilities. Here, the probability that E. coli is not found in one swimming area is 0.995. To find the combined probability that E. coli is not found in any of the 10 swimming areas, we multiply the probability for each sample together: \(0.995 \times 0.995 \times... \times 0.995\). This can be simplified using the exponent notation: \(0.995^{10} \). By doing this, we determine the likelihood of an event affecting all chosen subjects collectively, which is a key factor in determining the necessity of further testing.
The Importance of Public Health Testing
Public health testing is vital for ensuring the safety and well-being of communities. Testing water for pathogens like E. coli can prevent outbreaks of diseases caused by contaminated water. Combining samples can reduce costs and resources needed for testing. However, it's crucial to understand the probabilities involved to decide when more detailed testing is necessary. If the probability of contamination in a combined sample is below a certain threshold, fewer resources are wasted on unnecessary tests. This balance leads to more efficient public health management.
Escherichia coli Detection in Water Samples
Escherichia coli (E. coli) is a bacterium commonly found in the intestines of humans and animals. While most strains are harmless, some can cause serious illness. Detecting E. coli in water indicates fecal contamination and a potential health risk. In our exercise, the Fairfield County Department of Public Health aims to combine and test water samples to make the process cost-effective. A detection probability of 0.005 for individual samples seems low, but combining samples significantly increases the chance of detecting at least one contaminated source. This method ensures that public health officials can monitor and respond to contamination risks effectively, keeping the community safe.

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