Given that out of 1000 individuals where 60 use a drug. I am given the probability of a

Karina Trujillo

Karina Trujillo

Answered question

2022-06-13

Given that out of 1000 individuals where 60 use a drug. I am given the probability of a false positive is 0.009 and the probability of a false negative is 0.10. I'm trying to find the true positive and true negative.
Let D be the even that the user uses a drug and D C the event the user is not a drug user. I know that P ( D ) = 0.06 and P ( D C ) = 0.994 as well as P ( + | D c ) = 0.009 and P ( | D ) = 0.10. So that means that P ( + | D C ) = 0.009 0.009 990 = 8.91 people. Also P ( | D ) = 0.10 60 .10 = 6 people. Does it follow that P ( | D C ) = 981.09 / 990 = 0.991 and P ( + | D ) = 54 / 60 = 0.90?

Answer & Explanation

Dustin Durham

Dustin Durham

Beginner2022-06-14Added 31 answers

You computed too much, but the numbers you got are correct.
With no computation, Pr ( | D c ) = 1 0.009 = 0.991 and Pr ( + | D ) = 1 0.10 = 0.9.
Some steps in your computation were technically not right. For example, it is not true that Pr ( | D ) = 6 (people). Probabilities are always between 0 and 1.
Now you need to solve the real problem, which presumably has to do with finding Pr ( D | + ). For this you will need the definition of conditional probability, or Bayes' Formula.

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