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The number of positive and negative drug test results of the subjects is provided.

The frequencies are categorized as true and false results.

For two events A and B, the probability of occurrence of only A, only B, or both is computed as shown below:

$P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{and}B\right)$

Here, the probability of occurrence of both A and B is given by $P\left(A\text{and}B\right)$.

The table below shows the number of subjects that fall into each category:

The total number of subjects is equal to 300.

Let E be the event of selecting a subject who tested positive.

Let F be the event of selecting a subject who did not use marijuana.

The number of subjects who tested positive is equal to 143.

The number of subjects who did not use marijuana is equal to:

$$\begin{gathered} \text{False}\text{Positive}+\text{True}\text{Negative}=24+154 \\ =178 \end{gathered}$$

So, the corresponding probabilities are:

$$\begin{gathered} P\left(E\right)=\frac{143}{300} \\ P\left(F\right)=\frac{178}{300} \end{gathered}$$

The number of subjects who tested positive but did not use marijuana is equal to the number of subjects who tested false positive. Thus, it is equal to 24.

$P\left(E\text{and}F\right)=\frac{24}{300}$

The probability of selecting a subject who tested positive or did not use marijuana is equal to:

$$\begin{gathered} P\left(E\text{or}F\right)=P\left(E\right)+P\left(F\right)-P\left(E\text{and}F\right) \\ =\frac{143}{300}+\frac{178}{300}-\frac{24}{300} \\ =\frac{297}{300} \\ =0.99 \end{gathered}$$

Therefore, the probability of selecting a subject who tested positive or did not use marijuana is equal to 0.99.