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Use the following results from a test for marijuana​ use, which is provided by a certain drug testing company. Among 149 subjects with positive test​ results, there are 20 false positive​ results; among 150 negative​ results, there are 2 false negative results. If one of the test subjects is randomly​ selected, find the probability that the subject tested negative or did not use marijuana.​ (Hint: Construct a​ table.)

The probability that a randomly selected subject tested negative or did not use marijuana is ____.

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Final answer:

To calculate the probability that a subject tested negative or did not use marijuana, we sum the true negatives, false negatives, and false positives and divide by the total number of subjects. The probability is approximately 0.5689.

Step-by-step explanation:

To find the probability that a randomly selected subject tested negative or did not use marijuana, we can construct a table based on the provided information:

OutcomeNumber of SubjectsTrue Positives (Marijuana users correctly identified)129False Positives (Non-users incorrectly identified as users)20True Negatives (Non-users correctly identified)148False Negatives (Marijuana users incorrectly identified as non-users)2

The total number of subjects is 149 with positive results and 150 with negative results, giving us 299 subjects in total. To find the probability of a subject testing negative or being a non-user, we can add the true negatives and the false negatives together, as they represent the subjects who tested negative. In addition, non-users also include those who had false positive results.

So, the total number of subjects who tested negative or were non-users is 148 (true negatives) + 2 (false negatives) + 20 (false positives) = 170. Therefore, the probability of a randomly selected subject testing negative or not using marijuana is 170/299.

Using a calculator:

Probability = 170 / 299 ≈ 0.5689

So the answer is: The probability that a randomly selected subject tested negative or did not use marijuana is approximately 0.5689.

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