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Hi everyone I would I really appreciate if anyone could help me out with this problem. I was discussing it with a friend and we disagreed on whether we needed to treat this as a conditional probability problem or whether we just needed to multiply 94% and 98% to get the answer. This is not for a class, I am just interested in the topic.

The probability that a patient has HIV is 0.001 and the diagnostic test for HIV can detect the virus with a probability of 0.98. Given that the chance of a false positive is 6%, what is the probability that a patient who has already tested positive really has HIV?

Thanks in advance

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    I'm confused. Could you clarify what you mean by the accuracy of the testing method?2012-09-05
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    This resembles an example very commonly used to illustrate _Bayes' formula_.2012-09-05
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    I have the same question as Ben. Does accuracy mean probability of test being positive given subject has HIV?2012-09-05
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    Sorry about the confusion. I edited it, I hope now it's OK.2012-09-05
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    What does "the chance of a false positive" mean? Intuitively (and considering how high the given number is) I would understand it to mean "the probability that a patient who has already tested positive _doesn't_ have HIV". But is that is what it means, then your answer is just $1-0.94=0.06$ and neither of the numbers $0.001$ or $0.98$ are relevant.2012-09-05
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    @HenningMakholm: I take the chance of a false positive to be the chance that a patient without HIV tests positive.2012-09-05
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    @RossMillikan: That would be my guess too -- except that when I wrote the comment, the question claimed this probability to be 96%, which would have been ridiculous.2012-09-05
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    One way of working through such problems is to create a grid with HIV+ yes or no; positive on test yes or no. Take 1000 people, or some other convenient number like 10000, and use the information you have to fill in the grid. I find this aids clarity of thought.2012-09-05

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A positive result can come in two ways: a patient with HIV and a correct result: $0.001 \cdot 0.98=0.00098$ of the population, or a patient without HIV and a false positive. The fraction of false positives is $0.999 \cdot 0.06 = 0.05994$. The false positives are then $\frac {0.05994}{0.0094}\approx 61.2$ times more than the infected patients.

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    Sorry about all the typos, I edited the question again. Thanks for your answer.2012-09-05
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    @Daniel: I have updated in light of the edit.2012-09-05
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This is a classic example for the illustration of Bayes' theorem. Let's first formulate the problem in formal terms. Let $D$ be the event that the person has the disease, then $D^c$ denotes the event that the person doesn't have the disease. Let $Y$ be the event that the test gives the positive result (person has the disease as per the test diagnostic) and $N$ be the event that the test gives the negative result.

Now let's write down the given information. $$P(D) = 0.001$$ $$P(Y|D) = 0.98 $$ $$P(Y|D^c) = 0.06 $$ We have to find $P(D|Y) $.

Now we'll use Bayes' theorem to find the required probability. $P(D|Y) = \frac{P(Y|D)P(D)}{P(Y)}$. $$P(Y) = P(Y \cap(D\cup D^c)) = P(Y\cap D) + P(Y\cap D^c) = P(Y|D)P(D) + P(Y|D^c)P(D^c) $$ as $D$ and $D^c$ are mutually exclusive events and together form a partition of the sample space. Using the given values, we have $$P(Y) = 0.98 \times 0.001 + 0.06 \times 0.999 = 0.06092$$

Therefore, $$P(Y|D) = \frac{0.98 \times 0.001}{0.06092} = 0.016$$

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    Read Wikipedia for Bayes' theorem and conditional probability.2012-09-05
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    It is correct and you are a positive person so you normally get (+)2012-09-05
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    This should really be the accepted answer.2018-08-23