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The probability that a water sample from public swimming area contains E.coli is equal to 0.005.

The number of pools sampled is 10.

The probability that an event occurs at least once is one minus the probability that the event does not occur at all. For any given event A, it has the following notation:

$P\left(Aoccurringatleastonce\right)=1-P\left(Anotoccurring\right)$

Let A be the event that a selected water sample has E.coli.

It has the following probability:

$P\left(A\right)=0.005$

Here, $\stackrel{炉}{A}$is the event that a selected water sample does not have E.coli.

It has the following probability:

$$\begin{gathered} P\left(\stackrel{炉}{A}\right)=1-0.005 \\ =0.995 \end{gathered}$$

The probability that out of 10 selected samples, none has E.coli is computed below:

$$\begin{gathered} P\left(\text{none}\text{has}\text{E.coli}\right)=P\left(\stackrel{炉}{A}\right)脳P\left(\stackrel{炉}{A}\right)脳...脳P\left(\stackrel{炉}{A}\right)\left(10\text{times}\right) \\ ={\left(0.995\right)}^{10} \\ =0.951 \end{gathered}$$

The probability that the combined sample of 10 water samples has E.coli is equal to the probability that at least one of the 10 samples has E.coli. Thus, it is calculated as follows:

$$\begin{gathered} P\left(\text{at}\text{least}\text{one}\text{has}\text{E.coli}\right)=1-P\left(\text{none}\text{has}\text{E.coli}\right) \\ =1-0.9511 \\ =0.0489 \end{gathered}$$

Therefore, the probability that the combined sample will have E.coli is equal to 0.0489.

The value of probability can range between 0 and 1, inclusive of both.

An event with a probability lesser than 0.05 can be considered rare or unusual. Thus, the probability of getting a positive result is low for the combined sample.

This implies that further testing of the samples is not necessary.