# 52 card deck

##### New Member
Ok, I have been trying to solve part d) in my homework for the past hour and then I turned to the web and still couldn't find an answer. This is an advanced probability and statistics class and I am willing to work hard but I don't know how to do everything. I have solved a, b, and c.
The actual question is: Suppose that you are dealt 3 cards from a standard deck of cards.
d) Find the probability that you are dealt 3 Kings, given you were dealt at least 2 Kings.

I came so close to the correct answer with this: (42C2 * 48C1)/ 52C3 = 0.01303 but it's off by a little bit.

#### S-C-3-1-3

##### New Member
Welcome to the forums!

$$P(\text{3 Kings}|\text{2 Kings})=\frac{P(\text{3 Kings})}{P(\text{2 Kings})}=\frac{{4\choose 3}}{{4\choose 2}{48\choose 1}}=\frac{4}{288}\approx \boxed{0.014}$$

The general formula for conditional probability is $$P(A|B)=\frac{P(AB)}{P(B)}$$, and in the numerator, we are basically looking for the probability that we have 2 and 3 Kings, or 3 Kings in total. Then in the denominator, we choose the probability we have only 2 Kings: 2 from the 4 Kings and the remaining card from the 48 other cards.

##### New Member
The conditional probability given that the person already had two Kings is what was getting me sidetracked. I am just seeing this notation and still trying to get it to sink in.
I normally say during the problem, OK the denominator of the probability is always the no. of selections being made choosen from the whole sample space. With more practice I will get better at recognizing which questions place conditions on the solution and which are just wasting words in order to confuse me. The denominator in this problem shows me that 4C2 is the choice, 2 Kings, these occupy two of the 3 cards. Then 48C1 =48, fills in the what's implied by the question which is the 3rd card will be a King, right?
This notation is exactly like what my book has and I want to make sure I know what you are saying about the numerator, so what I think you mean that the numerator is the probability of having 2 Kings and the probability of having 3 Kings. So, 4C3 takes care of having 2 Kings.

#### BGM

##### TS Contributor
Yes the numerator is the intersection of two events. Note that

$$A \cap B = A$$ if $$A \subseteq B$$

And you can see the event of having 3 kings is a subset of the event of having 2 kings
(the former implies the latter)