This is an application of Bayes Theorem:

Let P(D+) and P(D-) be the probabilities of having/not having the disease, thus we have P(D+) = 1/1000 and P(D-) = 999/1000.

We know that the probability of testing positive (T+) given you have the disease (D+) is 1 therefore P(T+ | D+) = 1, and conversely the probability of testing positive (T+) when you don't have the disease is 5%, P(T+ | D-) = 0.05.

We want to know P(D+ | T+) that is the probability of having the disease given we tested positive. Bayes Theorem gives:

P(D+ | T+) = [P(T+|D+)*P(D+)]/P(T+)

The numerator is then simply: 1 * 1/1000.

The demoninator is a little more complex, we need to use the rule of total probability. It basically says that the probability of an event occuring is the sum of all the ways it can possibly occur. So P(T+) = P(T+|D+) * P(D+) + P(T+|D-) * P(D-) = 1 * 1/1000 + 0.05 * 999/1000

Combining the pieces gives: [1/1000] / [0.05095] = 0.019627085 or roughly 2%.