Q:

Probability of drawing three cards, with the third being a King

CONTEXT: First year university statistics course exercise
If $C$ is the number of cards drawn from a deck of 52 cards until a King
is drawn, find $\mathbb P(C = 3)$. I'm not 100% sure how to do this question. 
This would mean the first and second cards are not Kings but the third is. 
Let $A$ be the event that you draw two non-Kings and then a King.
$|A| = {48 \choose 1}{47 \choose 1}{4 \choose 1}$.
The sample space $\Omega$ would be any three card selections. $|\Omega|={52 \choose 1}{51 \choose 1}{50 \choose 1}$
By the equilikely principle, this means $\mathbb P(C = 3)=\frac{{48 \choose 1}{47 \choose 1}{4 \choose 1}}{{52 \choose 1}{51 \choose 1}{50 \choose 1}}$?
Any help would be greatly appreciated (I'm finding it difficult to check my counting answers!)

A:

I think your answer and reasoning is correct, but sometimes it's easier to think about the events separately; you have three independent events, P(not King 1st draw), P(not King 2nd draw | no King on 1st) and then P(King | no King on 1st or 2nd draws) and the final probability is just these multiplied together. 
$$\frac{48\choose{1}}{52\choose{1}}\times\frac{47\choose{1}}{51\choose{1}}\times\frac{4\choose{1}}{50\choose{1}}$$
I know it's equivalent but I find it helps even just thinking about it in a slightly different way.

