Well, first, if 1% of the blood samples really are from people with AIDS, then you can put 1% of 10,000 = 100 as the margin total in the first column, and that will leave 10,000 - 100 = 9,900 blood samples that don't have AIDS as the margin total in the second column. Next, if 99% of the people with AIDS will be correctly spotted by ELISA, 99 of the 100 samples in the first column will fall into the first row in that column, leaving 100 - 99 = 1 in the second row in that column. Next, if 94% of the 9900 people who don't have AIDS will be correctly told they don't have it by ELISA, I guess there should be .94 * 9900 = 9306 blood samples in the second row of the second column, leaving 9900 - 9307 = 693 for the first row of that column. Finally, then, the row margin totals are 99 + 594 = 693 and 1 + 9306 = 9307, the table is complete, and we can work out the conditional probability we want: P(person has AIDS given ELISA positive) = 99/693 = 14%. In other words, only about 14% of the people ELISA says have AIDS actually will have it! This seems like a disappointingly low figure given ELISA's apparently good performance numbers (99% and 94%), so it's worth taking a moment to see why.
The Truth
person | person does
has AIDS | not have AIDS
------------------------------------------------------
What ELISA ELISA positive | 99 | 591 | 693
says ELISA negative | 1 | 9,306 | 9,307
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100 9,900 | 10,000
There are a variety of points that can be made on the way to an explanation: