What is the probability that a patient who has already tested positive really has HIV? 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
 A: 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, $$P(Y|D) = \frac{0.98 \times 0.001}{0.06092} = 0.016$$
A: 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.
