I have only looked at problem 2a so far.

You have 18 Americans, 12 Foreigners.

1 of those Foreigners is chosen (striker). You now randomly select 3 more people. What are the chances that all three are American?

So now you have 18 Americans, 11 Foreigners.

the chance of first being an American is 18/(18+11) = 18/29

the chance of the second being an American is 17/(17+11) = 17/28

the chance of the third being an American is 16/(16+11) = 16/27

The chances of those happening consecutively?

(18/29)*(17/28)*(16/27) = (4896/21924) = 22.31%

Proof? I just got done making a simulation that would choose 3 players from 29 total. 18 of them are americans, 12 foreigners. Once one is chosen, it cant be chosen again.

Here were the results of the first test with 100,000 trials.

which is 22.188%

and here's the code.