Given two machines break down, what is probability one is from each company? A factory has four machines, of which two are imported from Country A and two are imported from Country B.
The probability that the machine imported from Country A breaks down is $0.3$.
The probability that the machine imported from Country B breaks down is $0.2$.
The event of breaking down for all four machines are mutually independent.
Given that exactly two machines in the factory broke down, what is the probability that these two machines consist of exactly one machine imported from Country A and one machine imported from Country B?
I tried to do something along the lines of ${2\choose1}(.3)(.7)+{2\choose1}(.2)(.8)$ but I don't think this is correct.
 A: What you tried to calculate is the probability of exactly one machine from each country breaking down, but that is indeed not what is being asked. they ask for the conditional probability that this is the case given that two machines break down.  More formally, if we let $AB$ be the event that one machine from each country breaks down, and $XX$ the event of two machines breaking down period, the question is to calculate $P(AB|XX)$, rather than $P(AB)$ which is what you tried to do.
Also, I say that you 'tried to calculate' $P(AB)$ since you didn't do this quite right: the probability of one machine from each country breaking down is $P(AB)=2*0.3*0.7\color{red}*2*0.2*0.8$ since all those events need to happen at the same time.
Anyway, besides calculating the chance of one machine breaking down from each country, you should also calculate the chance of two machines from A breaking down while both machines from B don't break down ( which is $P(AA)=0.3*0.3*0.8*0.8$) and vice versa (which is $P(BB)=0.7*0.7*0.2*0.2$). 
Now add up those 3 probabilities, and you have the chance that two machines break down, period. That is, $P(XX) = P(AB) +P(AA) +P(BB)$. So, if you then take the chance of exactly one from each country breaking down and divide it by the chance of two machines breaking down period, you have the conditional probability they are looking for. That is: $P(AB|XX) = \frac{P(AB)}{P(XX)}$
