What is the closed formula for the n-dimensional combinatoric problem below? (probability,combination)?

Question

Suppose you have an examination with R questions on it. Suppose each question requires you to know F pieces of information in order to get it right. (and there are F pieces of info for each question, i.e. you need to get all the parts right to get the question right)

Can you write a closed formula to calculate the probability of getting C questions correct if you know L pieces of information out of the total F*R?
Or, can you write a closed formula to calculate the probability of having known L pieces of information given that you got C questions correct? (then you can flip it with Bayes' Theorem to the first one)

The answer must be valid for F and R being any positive integers. I'd also need to know why the formula works. Thanks in advance!

Stym, you almost have it; that's the point that I'm at as well. The one thing that your formula doesn't account for is the fact that when you multiply COMBIN(R,C) with COMBIN(RF-FC,L-FC) and L-FC>=F, you duplicate some possibilities. For example, if R=3, F=2, and L=4, you would expect the probability of getting at least 1 correct to be 1, as you are guaranteed to get at least 1 right. The formula, however, gives you 1.2, an impossible probability. Try plugging in R=3, F=2, and L=6, and see what the probability of getting 0 correct is. It should be 0, but it comes out as a negative integer! Any further ideas you are anyone else have on this would be great; and even if you don't know how to deal with this (outside of a recursive summation), thanks for taking time to think about it.

Answer 1

Represent the information in a matrix with R lines and F columns. Each piece of info you know is a 1 (there is no rule spcifying you are more likely to know certain pieces of information more than the others) and each piece of info you don't know is a 0.
Now we can work.
You have L pieces of information to place in RF spots. That gives you a total of Comb(RF,L) possible arrangements. That is (RF)!/(L!(RF-L)!)

Now let us compute the possible arrangements that yield at least C correct answers.
You will have C lines in your matrix that are all ones, representing the C questions you can answer. There are Comb(R,C) possible ways to choose those C lines in your matrix, or R!/(C!(R-C)!)
You are left with L-FC ones to place in the R-C lines left.
R-C lines is RF-RC spots and you then have Comb(RF-RC,L-FC) ways to place the remaining ones.
This supposes that L≥FC. Of course, if L Note also that the remaining arrangements don't exclude the possibility of another full line of ones. This means that I am calculating the probability to get at least C correct answers.

That probability is Comb(R, C)*Comb(RF-RC, L-FC)/Comb(RF, L) which is the number of favorable arrangements divided by the total possible arrangements.

P(at least C) = R!/(C!(R-C)!)*(RF- FC)!(/(RF-L)!(L- FC)!)/(RF)!*L!(RF- L)!
P(at least C) = R!L!(RF- FC)!/(RF!C!(R-C)!(L- CF)!)

The probability to get EXACTLY C answers is P(at least C) - P(at least C+1). You can write it using our formula only if L ≥ FC+F
P(exactly C| L≥FC+C) = R!L!(RF- FC)!/(RF!C!(R- C)!(L-CF)!)
- R!L!(RF- FC-F)!/(RF!(C+1)!(R- C-1)!(L-CF -F)!)

if FC≤ L < FC+F, your probability is only our first formula (because P(at least C+1) is exactly 0) and then P(exactly C) = P(at least C)

finally if L ≤ FC, P(exactly C) = 0