I am a bit confused about this problem. I understand that you need to pick a sample first, K, and then find the probability of that sample being red, L. The total different combinations of picking a sample to begin with is C(K,N). The total combinations of picking a red from the sample is C(L,K), however I think I am missing something here. Can someone please help explain this to me? Much appreciated.
C(selected number,total number) is the notation for combinatorics
There are N socks in a drawer, out of which M are red. Let us say you select K socks at random, without placing any back after the selection. What is the probability that L socks out of K selected are red, assuming the probability of any given outcome of this experiment is the same?
In this answer your $C(\text{selected},\text{total})$ is denoted as $\binom{\text{total}}{\text{selected}}$
Picking $K$ socks out of $N$ can be done on $\binom{N}{K}$ ways.
Picking $L$ red socks out of $M$ can be done on $\binom{M}{L}$ ways.
Picking $K-L$ not red socks out $N-M$ can be done on $\binom{N-M}{K-L}$ ways.
This leads to a probability of: $$\frac{\binom{M}{L}\binom{N-M}{K-L}}{\binom{N}{K}}$$
addendum
Another approach/way of thinking:
There are $K$ elected socks and $N-K$ not elected socks.
The socks are ordered in such a way that $L$ of the elected socks and $M-L$ of the non-elected socks have color red.
This can be done on $\binom{K}{L}\binom{N-K}{M-L}$ ways.
Ordering $N$ socks of which $M$ have red color, can be done on $\binom{N}{M}$ ways.
So this amounts in a probability of: $$\frac{\binom{K}{L}\binom{N-K}{M-L}}{\binom{N}{M}}$$
In fact we have:
$$\frac{\binom{\text{red}}{\text{red and selected}}\binom{\text{not red}}{\text{not red and selected}}}{\binom{\text{total}}{\text{selected}}}=\frac{\binom{M}{L}\binom{N-M}{K-L}}{\binom{N}{K}}$$ $$=\frac{\binom{K}{L}\binom{N-K}{M-L}}{\binom{N}{M}}=\frac{\binom{\text{selected}}{\text{red and selected}}\binom{\text{not selected}}{\text{red and not selectd}}}{\binom{\text{total}}{\text{red}}}$$