probability help

A manager wants to start drug testing at work, and he wants to find out how accurate the tests truly are. Suppose a drug tests 99% accurate. This means, the probability that the test is positive given that the person does drugs is 99%. Likewise, the probability that the test is negative given that the person is a non drug user is 99% also. The probably that people do drugs in the workforce is .5%. Using Bayes’ Theorem find the probabilities of the next two questions.

What is the probability that an employee is a drug user given the test is positive?



What is the probability that an employee is a drug user given that the test is negative?

las123 said:

A manager wants to start drug testing at work, and he wants to find out how accurate the tests truly are. Suppose a drug tests 99% accurate. This means, the probability that the test is positive given that the person does drugs is 99%. Likewise, the probability that the test is negative given that the person is a non drug user is 99% also. The probably that people do drugs in the workforce is .5%. Using Bayes’ Theorem find the probabilities of the next two questions.
What is the probability that an employee is a drug user given the test is positive?

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Use the notation \(\displaystyle D\) for the event is a drug user and \(\displaystyle \overline{D}\) for is not drug user.rom
From the given we know that \(\displaystyle P(+|D)=0.99~P(-|\overline{D})=0.99~\&P(D)=0.005\)

From the above we know that \(\displaystyle P(-|D)=0.01~\&~P(+|\overline{D})=0.01\)

To answer that question we need \(\displaystyle P(+)=P(+|D)P(D)+P(+|\overline{D})P(\overline{D})\)

If you can work that out, then answer the question \(\displaystyle P(D|+)\).

Actually, it would be unusual for the percentages of false positives and false negatives to be the same!

However, here is how I would do it. Assume, for simplicity, a population of 10000 people. Since ".5% do drugs", there will be .005(10000)= 500 people who use drugs and 9500 who do not. Of the 500 people who use drugs, 99%, .99(500)= 495 will test positive, 5 will not (false negatives). Of the 9500 people who do not do drugs, 1%, .01(9500)= 95 will test positive (false positives), 9405 will not. That is, of the 495+ 95= 590 people who tested positive, 495 or 495/590= .838 or about 84% actually use drugs.

HallsofIvy said:

However, here is how I would do it. Assume, for simplicity, a population of 10000 people. Since ".5% do drugs", there will be .005(10000)= 500 people who use drugs and 9500 who do not. Of the 500 people who use drugs, 99%, .99(500)= 495 will test positive, 5 will not (false negatives). Of the 9500 people who do not do drugs, 1%, .01(9500)= 95 will test positive (false positives), 9405 will not. That is, of the 495+ 95= 590 people who tested positive, 495 or 495/590= .838 or about 84% actually use drugs.

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Surely .005(10000)= 50.