We are conducting a test on a rare disease. A positive result means that, according to the test, the subject is infected. The following characteristics are known about the test and the disease: If a person is infected, the person has a $95\%$ chance of testing positive. When a healthy person is tested, the test has a $99\%$ chance of giving a negative result. A mere $0.1\%$ of the population is infected with the disease. If a person is tested positive for the disease. What is the probability that the person actually has the disease? I don't understand how to write the conditional probabilities if someone can explain to me than I should know how to solve it, but I'm not sure how to convert the text into the mathematics language.


closed as off-topic by Xander Henderson, mechanodroid, Did, BruceET, Jam Sep 16 '18 at 23:33

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  • $\begingroup$ I'm struggling to get that answer. I honestly just don't know how to approach this question. $\endgroup$ – FTAC Sep 16 '18 at 21:01
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    $\begingroup$ This is a fairly standard exercise in elementary prob and stats. Indeed, it is one of the questions that I put on my intro stats exam last fall. Our job here is not to answer your standard exercises for you, but to answer questions about mathematics. If you just need an answer to this standard exercise, there are other resources out there (almost any intro stats book should have a solution). If you read a standard answer and there is some particular aspect of one of that solution which you don't understand, then you should ask about that confusion here. $\endgroup$ – Xander Henderson Sep 16 '18 at 21:01
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    $\begingroup$ Hi. If you want an answer, then you should show your efforts. (No work is a standard problem with new posts, I learned this the hard way) $\endgroup$ – Jason Kim Sep 16 '18 at 21:03
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    $\begingroup$ As an exercise, I Googled the exact title of your post: "What is the probability that a person actually has the disease?." This gets me a ton of answers. One of these is likely to prove helpful to you. $\endgroup$ – Xander Henderson Sep 16 '18 at 21:05
  • $\begingroup$ if T is my positive test probability and I is my infected probality, I wrote P(T|I)=0.95, P(I)=0.1, P(Tcomp|Icomp)=0.99 and I should finf P(I|T), right? $\endgroup$ – FTAC Sep 16 '18 at 21:10

This involves using the Law of Total Probability and Bayes' Theorem. Let + and - represent positive and negative tests; let D and N represent having the disease and not. Here is an outline to get you started. Try to match each part to formulas in your text.

You seek $P(D|+) = P(D \cap +)/P(+).$

For the numerator, use $P(D \cap +) = P(D)P(+|D),$ where numerical values for both factors are given in the statement of the problem.

For the denominator, start with $P(+) = P(D \cap +) + P(N \cap +).$ You already know the first term from the numerator. Find a similar way to evaluate the second term from information given in the problem.

  • $\begingroup$ Now my question is, how can I calculate the probabilities in the debìnominator? $\endgroup$ – FTAC Sep 16 '18 at 21:31
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    $\begingroup$ Now my question is why you don't show what you tried in that direction. $\endgroup$ – BruceET Sep 16 '18 at 21:44
  • $\begingroup$ When a healthy person is tested, the test has a 99% chance of giving a negative result. Should be P(-|N), right? How can I relate it with P(N∩+)? $\endgroup$ – FTAC Sep 16 '18 at 21:49
  • $\begingroup$ Similar to $P(D \cap +) = P(D)P(+|D),$ except knowing $P(-|N)$ you need to be able to deduce $P(+|N).$ $\endgroup$ – BruceET Sep 18 '18 at 3:24
  • $\begingroup$ Solved, thank you! $\endgroup$ – FTAC Sep 18 '18 at 3:28

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