Probability of choosing three cards of same suit with 1 king I'm having a little trouble wrapping my head around conditional probability so I'm doing some exercises to try to better understand. This problem came up:
Three cards are randomly selected from a fair deck (without replacement).      
Find the probabilities of the following events by applying the definition 
of conditional probability:

• All hearts, given all cards are red.
• All hearts, given one of the cards is a king
• All hearts, given no spades.
• All hearts, given two kings

The first one makes sense because hearts are already red, so I multiply each selection like this to get the probability of three hearts
${1\over4} * {12\over51} * {11\over50} = {11\over850}$
and like this to get probability of three red cards
${1\over2} * {25\over51} * {24\over50} = {2\over17}$
By the theory of conditional probability
$P(A|B) = {P(A∩B)\over P(B)}$
I get ${11\over 850} \over {2\over 17}$ which is roughly 11.7% (Did I do this right?)
For the next case, I'm a bit stumped on how to proceed. There is only one king of hearts, so the odds of drawing it is ${1\over52}$ from a deck with no cards already drawn. But how do I take into account that the king might be the second or third card that I draw?
Any help is much appreciated.
 A: Since the order the cards are dealt have no importance, you should work with combinations instead of permutations. Therefore, if we set $A$="Hands with 3 hearts" and $B$="Hands with 3 red cards", we get
$$
P(A \cap B) = \frac{{13 \choose 3}}{{52 \choose 3}} \quad, \quad P(B) = \frac{{26 \choose 3}}{{52 \choose 3}}
$$
and therefore
$$
P(A | B) = \frac{{13 \choose 3}}{{26 \choose 3}}
$$
For the next case we set $A$="Hands with 3 hearts" and $B$="One card is a king".
$A \cap B$ are hands with $3$ hearts, one being a king (The King of hearts necessarily). Therefore there are ${12 \choose 2}$ such hands (The King is already determined). Also, $|B|={52 \choose 3}-{48 \choose 3}$ (All possible hands minus those without a king). We then get
$$
P(A \cap B) = \frac{{12 \choose 2}}{{52 \choose 3}} \quad, \quad P(B) = \frac{{52 \choose 3}-{48 \choose 3}}{{52 \choose 3}}
$$
therefore
$$
P(A | B) = \frac{{12 \choose 2}}{{52 \choose 3}-{48 \choose 3}}
$$
The same idea applies for the next two problems.
Hope this helps.
