To compute this probability, we will have to compute how many outcomes are possible for five draws from a deck and how many of these have exactly 3 aces and 2 kings.
Assuming it is a standard deck of 52 cards, we have a total of 4 aces and 4 kings in the deck.
This means that we need to calculate the combinations in two groups and then combine these groups.
The first group are the 3 aces. Since the order doesn't matter, we have a case of "4 choose 3":
[tex]C_1=\frac{4!}{3!(4-3)!}=\frac{4!}{3!}=4_{}[/tex]The other group are the kings, but we will pick only 2, so it is "4 choose 2":
[tex]C_2=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=3\cdot2=6[/tex]Now, we want to combine these two groups, we do it by multiplying their possible combinations:
[tex]C=4\cdot6=24[/tex]So, there are 12 possible hands with 3 aces and 2 kings.
Now, we ned to compute the total possible outcomes. Since we have a deck of 52 cards and will pick 5, this is "52 choose 5":
[tex]C_a=\frac{52!}{5!(52-5)!}=\frac{52!}{5!47!}=\frac{52\cdot51\cdot50\cdot49\cdot48}{5\cdot4\cdot3\cdot2\cdot1}=\frac{52\cdot17\cdot5\cdot49\cdot12}{1\cdot1\cdot1\cdot1\cdot1}=2,598,960[/tex]Then the probability will be the combinations of the hand we want over the total combinations:
[tex]P=\frac{24}{2,598,960}=\frac{1}{108,290}\approx0.00000923446301597562[/tex]