False positive

I will use C to mean that a randomly selected individual is infected, $\bar{\text{C}}$ to mean that they do not have the infection and "positive" to mean that the test indicates that the person has the disease.

I will use P(A|B) to mean the probability that A is true given that B is true. So we want to find P(C|pos)

We can use the multiplicative rule for conditional probabilities to find P(C and pos) in two ways

P(C and pos) = P(C) P(pos|C) = P(pos) P(C|pos)

and so

$P(C|pos)=\dfrac{P(pos|C)P(C)}{P(pos)}\dots(1)$

The denominator can be written as the sum of the probabilities of two mutually exclusive possibilities - either the selected individual is infected and the test gives the correct result, or the person is not infected and the test gives a so called false positive. So (1) can be written as

$P(C|pos)=\dfrac{P(pos|C)P(C)}{P(C)P(pos|C)+P(\bar{C})P(pos|\bar{C})}$

Finally put in the numbers from the question to get

$P(C|pos)=\dfrac{0.99\times10^{-4}}{0.99\times 10^{-4}+0.05\times 0.999}\approx \boxed{0.002}$

Assume we test $10^6$ people.

Hence, $99$ people are tested true positive and $49995$ are tested false positive.

Therefore, the required probability is: $\frac {99} {49995+99}= 0.00198\approx 0.002$

This is an illustration of the False positive paradox

p(negative | not affected ) = 0.95.

Thus, p(positive | not affected ) = 0.05.

p(positive & not affected) = p(positive | not affected) * p(not affected)

= 0.05 * 0.9999 = 0.049995.

p(positive & affected) = p(positive | affected) * p(affected)

= 0.99 * 0.0001 = 0.000099.

p(positive) = p(positive & (not affected or affected))

= 0.049995 + 0.000099 = 0.050094.

Bayes' rule:

p(affected | positive) = p( positive | affected ) * p( affected ) / p(positive)

= 0.99 * 0.0001 / 0.050094 = 0.00197628458