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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 0.049, which is low enough to rarely require further testing.

Step by step solution

01

Define the Probability for one sample

The problem states that there is a probability of 0.005 of finding E. coli in a single swimming area. This probability can be denoted as: \[ P(E) = 0.005 \]
02

Calculate the Probability of No E. coli in one sample

The probability of not finding E. coli in one sample is:\[ P(\text{No E}) = 1 - P(E) = 1 - 0.005 = 0.995 \]
03

Find the Probability of No E. coli in 10 samples

For 10 independent samples, the probability that none of them contain E. coli is:\[ P(\text{No E in 10}) = P(\text{No E})^{10} = 0.995^{10} \]
04

Calculate the Probability That At Least One Sample Contains E. coli

The probability that at least one sample contains E. coli in 10 samples is:\[ P(\text{At least one E}) = 1 - P(\text{No E in 10}) = 1 - 0.995^{10} \]
05

Compute the Result

Calculate the value:\[ P(\text{At least one E}) = 1 - 0.995^{10} \]Using a calculator, this is approximately:\[ P(\text{At least one E}) \approx 1 - 0.951 = 0.049 \]
06

Determine if the Probability is Low

A probability of 0.049 (or 4.9%) is obtained. Since this value is less than 5%, it is considered low. Therefore, further testing of the individual samples would rarely be necessary if a combined sample tests positive.

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Key Concepts

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

probability calculation
Probability helps us understand how likely an event is to happen. In the context of composite water sampling, we want to figure out the chance of finding E. coli in combined water samples.

The problem starts by telling us the probability of finding E. coli in a single sample is 0.005. This is written as
\( P(E) = 0.005 \)

Then, we find the probability of not finding E. coli in a single sample, which is the complement of finding it. This probability is: \(
P(\text{No }E) = 1- P(E) = 0.995 \)

Next, to find the probability that none of the 10 samples have E. coli, we use:
\( P(\text{No }E \text{ in } 10 ) = 0.995^{10} \)

Finally, the probability that at least one of the 10 samples contains E. coli is: \( P(\text{At least one }E ) = 1 - 0.995^{10} \)

Doing the math, we get:
\( P(\text{At least one }E) = 1 -0.951 = 0.049 \)

This means there is a 4.9% chance that at least one of the 10 samples will have E. coli.
E. coli testing
Testing for E. coli in water is crucial for public health. E. coli is a type of bacteria commonly found in the intestines of humans and animals. Most strains are harmless, but some can cause serious illnesses.

When testing water, particularly in public swimming areas, detecting E. coli indicates that the water might be contaminated with fecal matter, which can carry pathogens.

To save costs, health departments like Fairfield County combine water samples from different locations. If this combined sample tests positive, they do further testing to identify the specific contaminated samples.

Combining samples is efficient, but it relies heavily on probability to minimize false negatives (missing contamination when it's there) and false positives (finding contamination that isn't actually there).

Thus, understanding and correctly calculating probabilities is essential in the process.
public health statistics
Public health statistics involve collecting and analyzing health data to make decisions that protect and improve community health. In our example, the statistics help to determine the likelihood of water contamination in public swimming areas.

Public health departments use this data to make informed decisions about testing policies, resource allocation, and public warnings.

By looking at historical data and probabilities, health officials can focus their efforts where risks are higher and ensure that the most effective preventative measures are in place.

These statistics are particularly vital in managing outbreaks, controlling the spread of diseases, and ensuring overall public safety.
independent events in probability
In probability, an independent event means that the occurrence of one event does not affect the occurrence of another. This concept is critical in our E. coli example.

Each water sample from the 10 different swimming areas is considered an independent event. The probability of finding E. coli in one sample doesn't affect the probability in another.

This independence allows us to calculate the combined probabilities accurately. For example, the likelihood that none of the 10 samples contain E. coli is calculated by raising the individual probability of no E. coli to the power of 10:
\( P(\text{No }E \text{ in } 10 ) = 0.995^{10} \)

Understanding independent events helps us accurately assess risks and make informed decisions about further testing and public health measures.

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