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