Conditional probability of an event given two independent events

#1
I am dealing with an interesting probability problem. I had 16 subjects randomly divided into 4 different rooms, each room having 4 seats. The subjects were given a question to solve. We knew from our prior experience that only 50% people could solve the problem. In our study, 7 of the 16 people solved and the others couldn't solve the problem. Both the solving times and the giving up times were recorded. So there were 7 ordered solving times and 9 ordered giving up times in our hands at the end of the study.

A subject is selected for the next task if he is the 1st solver of the 1st room or the 2nd solver of the 2nd room or the 1st solver of the 3rd room or the 2nd solver of the last room. So, it constitutes something close to a ranked set sample for the next task.

I want to find out the probability that the subject having the 5th ordered solving time was the 1st solver of the 1st room or the 2nd solver of the 2nd room or the 1st solver of the 3rd room or the 2nd solver of the last room. That is, I want to find out the probability of the 5th ordered solver to be selected for the next task.

Now, for being selected for the next task he should be able to solve the problem and the room he belongs to should have at least the necessary number of solvers. Say, a subject was randomly assigned to room 4, solved the problem, but room 4 had only 1 solver, then he could not be selected for the next task. So, the probability for ordered subject i seems like,

P(i gets selected for the next task|i solves the problem, the room it falls into has at least necessary number of solvers).

The probability seems to me like P(A|B,C) where events B and C are independent. How do I accurately calculate the probability? It seems complicated.

Thanks in advance for any help!
 
#3
Because we need to know if solving the next task has any relationship like only the quicker solvers of the first problem can solve the second problem, or can solve the next problem quicker again!