Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as showing HIV+ive. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV+ive. What is the probability that the person actually has HIV?
To determine the probability that the person actually has HIV given that they tested positive, we can use Bayes' theorem. Bayes' theorem allows us to update our initial knowledge about a situation based on new evidence.
Let's define the events:
A = Person has HIV
B = Person tests positive for HIV
We are interested in finding P(A|B), which is the probability that the person actually has HIV given that they tested positive.
According to the problem statement, we are given the following probabilities:
P(A) = 0.001 (0.1% of the population has HIV)
P(B|A) = 0.9 (90% of people with HIV test positive)
P(B|A') = 0.01 (1% of people without HIV test positive)
Now, let's calculate the probability using Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
To calculate P(B), we need to consider the probabilities of testing positive in both the HIV-positive and HIV-negative populations:
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
P(A') represents the complement of A, which means a person does not have HIV.
We can substitute the given values in the equation:
P(B) = (0.9 * 0.001) + (0.01 * (1 - 0.001))
P(B) = 0.0009 + 0.00999
P(B) = 0.01089
Now, we can calculate P(A|B):
P(A|B) = (0.9 * 0.001) / 0.01089
P(A|B) = 0.0009 / 0.01089
P(A|B) ≈ 0.0827
Therefore, the probability that the person actually has HIV given that they tested positive is approximately 0.0827, or about 8.27%.