Probability of getting exactly one ace and exactly one king in your opening hand. In a 52 cards deck, what is the probability of getting exactly one ace and exactly one king in your openning hand (5 cards)?
At first I was calculating the odds like this:
5C2 (44/52) (43/51) (42/50) (4/49) (4/48) = 0.04
But then I realized I have 3 types of cards, aces, kings and non-aces/non-kings. So I'm not sure I can use the binomial factor in this exercise.
Thanks
 A: Answer:
Exactly 1 Ace could be picked in ${4\choose1}$ ways, similarly exactly 1 King could be picked in ${4\choose1}$.  Now you have 8 cards removed from the deck and the remainder 3 cards should come from 44 cards which are non - ace and non-king.  That you can choose in ${44\choose3}$.  Lastly, 5 cards could be chosen in ${52\choose5}$
Thus the probability is $$\approx \dfrac{{4\choose1}.{4\choose1}.{44\choose3}}{{52\choose5}}$$
EDIT:
Other way is :
The Ace could be chosen with a probability $= \frac{4}{52}$
The King could be chosen with a probability $= \frac{4}{51}$
Rest of the three cards could be chosen with a probability  $= \frac{44}{50}.\frac{43}{49}
.\frac{42}{48}$
These five cards could be picked in $=\frac{5!}{1!.1!.3!}  $
Now multiply $$=\frac{4}{52}.\frac{4}{51}.\frac{44}{50}.\frac{43}{49}
.\frac{42}{48}.\frac{5!}{1!.1!.3!}$$
The first method is an approximate where as the second one is the most accurate.
A: This is what is called a hypergeometric distribution (drawing w/o replacement) 
I would prefer to solve it using combinations for choosing from the $3$ distinct groups
$Pr = \dfrac{\binom41\binom41\binom{48}3}{\binom{52}5}$
Your attempt will give the same answer if you use the appropriate multinomial coefficient as noted against your question. 
