Task:
A test for COVID-19 desease has probability for positive result from infected person equal to 0.9 and probability for negative result from noninfected person equal to 0.8. Half of the population is affected. What is the probability for a person to be noninfected if he has two negative tests? The tests are independent from one another.
Solution:
Let:
P - positive result
P' - negative result
I - infected person
I' - noninfected person
$P(P|I)=0.9$
$P(P'|I)=0.1$
$P(P'|I')=0.8$
$P(I)=0.5 => P(I')=0.5$
Probability for one negative test - using Bayes' theorem:
$$P(I'|P') = \frac{P(P'|I').P(I')}{P(P'|I').P(I') + P(P'|I).P(I)}$$ $$P(I'|P') = \frac{0,8.0,5}{0,8.0,5 + 0,1.0,5} = \frac{0,4}{0,4+0,05} = \frac{0,4}{0,45} = 0,8(8) ~= 0,89$$
How can I evaluate the probability of the person NOT being infected if he has TWO NEGATIVE tests?
Firstly, as @lulu noted, you should add all of your assumption to your problem, not to solution!
Secondly, by assumption of independency: $$ Prob(\text{NonInfected | 2 Negative}) = 1 - Prob((\text{NonInfected | 2 Negative})^c) = $$ $$ = 1 - Prob(\text{Infected | 2 Negative}) = 1 - (Prob(\text{Infected | Negative}))^2 = $$ $$ = 1 - (0.89)^2 = 1 - 0.7921 = .02079 $$