What is the probability that an enlarged hand contains at least $3$ kings

Question

Question: A card hand selected from a standard deck consists of $2$ kings, $1$ queen, $1$ jack, and $1$ ten. Three additional cards are selected at random and without replacement from the remaining cards in the deck. What is the probability that the enlarged hand contains at least $3$ kings?

There are $47$ cards remaining in the deck. Two are kings and $45$ are non-kings. My thinking is: if two kings are left in the deck, how am I supposed to get $3$? This looks like a binomial coefficient problem in my opinion. Need help.

Answer 1

You obtain three or more kings so long as you don't draw zero kings, so the answer is

$$1-P(\text{zero kings})=1-\frac{45}{47}\cdot\frac{44}{46}\cdot\frac{43}{45}=\frac{135}{1081}\approx 0.125$$