Conditional probability: At least 3 kings given there are at least 2 kings in the hand of 13. My first "conditional probability" problem. Sorry for all the questions. My instructor doesn't make sense to the class.
A hand of 13 cards is to be dealt at random and without any replacement from an ordinary deck of playing cards. 
Find the conditional probability that there are at least three kings in the hand given that the hand contains at least two kings.
A deep, good, and thorough explanation would be tremendously appreciated. 
Thanks
 A: There are $\binom{52}{13}$ ways to select 13 cards from 52 cards.
Consider how you would create a hand of 13 cards with the required number of kings.
There are $\binom{4}{n}$ ways of choosing the required number of kings.
There are then $\binom{48}{13-n}$ ways of choosing the remaining (13 - n) cards.  
Let A be the event of getting at least 2 kings.
This can be thought of as 3 separate events corresponding to 2, 3 or 4 kings.
By the addition principle:   
Number of ways of getting A = $\binom{4}{2}$ * $\binom{48}{13-2}$ + $\binom{4}{3}$ * $\binom{48}{13-3}$  + $\binom{4}{4}$ * $\binom{48}{13-4}$      
So, $$P(A) = \frac{\binom{4}{2} * \binom{48}{13-2} + \binom{4}{3} * \binom{48}{13-3}  + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Let B be the event of getting at least 3 kings.
This can be thought of as 2 separate events corresponding to 3 or 4 kings.
By the addition principle:
$$P(B) = \frac{\binom{4}{3} * \binom{48}{13-3}  + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Finally, by the conditional probability formula:   
$$P(B|A) = \frac{P(B \cap A)}{P(A)}= \frac{P(B)}{P(A)}$$
A: My Probability is not so good but still here's a solution.
Let $\Pr(A)$ is the probability of having at least two King cards in a hand of 13 cards. 
Hence, $\Pr(A) = \binom{4}{2} * \binom{48}{11} + \binom{4}{3}*\binom{48}{10}+\binom{4}{4}*\binom{48}{9}$
And let $\Pr(B)$ is the probability of having at least three King cards in a hand of 13 cards.
And let $\Pr(C)$ is the probability of "having at least 2 king cards and having at least 3 kings cards" simultaneously. This is indeed the case you have at least 3 king cards in one hand of 13 cards. 
Basically, $\Pr(C)=\Pr(A \cap B)$
Hence, $ \Pr( C )= \binom{4}{4}*\binom{48}{9}+ \binom{4}{3}*\binom{48}{10}$
So, required probability is $\Pr (B|A)= \large\frac{\Pr(C)}{\Pr(A)}$ 
I have used the conditional probability formula.
