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The probability that a selected sample test positive for drug use is 0.126.

The number of subjects selected is 5.

The probability that an event occurs at least once is one minus the probability that the event does not occur at all.

Mathematically, if A is an event,

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

In this case, it is required to compute the probability that the five samples combined test positive.

Assume that a combined sample will test positive if at least one of the sample results is positive.

Let A be the event that a selected sample is positive.

It has the following probability:

$P\left(A\right)=0.126$

Here, $\stackrel{炉}{A}$is the event that a selected sample is negative.

It has the following probability:

$$\begin{gathered} P\left(\stackrel{炉}{A}\right)=1-0.126 \\ =0.874 \end{gathered}$$

The probability that out of five selected samples, none is positive is computed below:

$$\begin{gathered} P\left(\text{none}\text{is}\text{positive}\right)=P\left(\stackrel{炉}{A}\right)脳P\left(\stackrel{炉}{A}\right)脳P\left(\stackrel{炉}{A}\right)脳P\left(\stackrel{炉}{A}\right)脳P\left(\stackrel{炉}{A}\right) \\ =0.874脳0.874脳0.874脳0.874脳0.874 \\ =0.510 \end{gathered}$$

The probability that the combined sample of five individual samples is positive is equal to the probability that at least one of the five samples is positive. Thus, it is calculated as follows:

$$\begin{gathered} P\left(\text{at}\text{least}\text{one}\text{is}\text{positive}\right)=1-P\left(\text{none}\text{is}\text{positive}\right) \\ =1-0.5100 \\ =0.490 \end{gathered}$$

Therefore, the probability that the combined sample will have a positive result is equal to 0.490.

When combined for five subjects, the chances that the combined sample tests positive is close to 0.5, which is moderate.

The resultant to this implies that further testing of the samples is necessary to detect the individual who is positive for drug use.