Counting probability of disease There are 2 types of people in a big factory: prone to COVID (30% of workers), and nonprone to COVID (70% of workers). For a COVID-prone person, p(COVID) = $0.4$. For a nonprone to COVID person, p(COVID) = $0.2$.
Compute:
a) the probability a random factory worker will have a COVID
b) given a person who has a COVID, what is p(person is COVID-prone).
I'm not sure, should I use binomial probability or usual, or how I should count in general. I suppose in (a) I need to calculate $0.3*0.4 + 0.7*0.2 = 0.26$

If we write everything we know, we got :
$P(Covid~\vert~prone~Covid) = 0.4 $
$P(prone~Covid) = 0.3$
$P(non-prone~Covid) = 0.7$
$P(Covid~\vert~non-prone~Covid) = 0.2$
We can deduce : $P(Covid) = P(Covid\bigcap prone~Covid) + P(Covid \bigcap non-prone~ Covid)$
This expression corresponds to the answer you found for a)
For b) : You have to calculate $P(prone~Covid \vert ~Covid)$ using what you found previously
