Problem Solving

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?

[(0.001)(0.9)]/[(0.999)(0.01)+(0.001)(0.9)] = 0.0826 or 8.26%. This will never be a GMAT question. Simply cannot do it without a calculator. The method behind this;

Truly HIV+/Total HIV +