Conditional Probability 5 card hand We have a 5 card hand from a standard deck. What is the probability that the hand is all Spades, given that it has at least two Spades? 
I know the formula for conditional probability is:
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
So in this case would it be:
$P(A|B) = \frac{13 \choose 5}{{13 \choose 5}+{13 \choose 4}{39 \choose 1}+{13 \choose 3}{39 \choose 2}+{13 \choose 2}{39 \choose 3}}$
 A: The number $X$ of spades can be modeled as a hypergeometric random variable.
$P(X = k) = \frac{{13\choose k}{39 \choose 5-k}}{{52 \choose 5}},$ for $k = 0,1,
\dots,5.$ Here is the PDF table from R statistical software:
k=0:5;  pdf = dhyper(k, 13, 39, 5)
cbind(k, pdf)
## k          pdf
## 0 0.2215336134
## 1 0.4114195678
## 2 0.2742797119
## 3 0.0815426170
## 4 0.0107292917
## 5 0.0004951981

I think your answer is OK, but I have not checked the numerical values directly from your formula.
If you're using a calculator to get numerical values, it is easier to find $P(X \ge 2) = 1 - P(X \le 1).$
From R, I get your desired conditional probability to be 0.001349141 (as Commented by @user1775500).
[In R, dhyper is the PDF and phyper is the CDF.]
dhyper(5, 13, 39, 5)/(1 - phyper(1, 13, 39, 5))
## 0.001349141
dhyper(5, 13, 39, 5)/sum(dhyper(2:5, 13, 39, 5))
## 0.001349141

It is extremely rare to get five spades in a five-card hand (.0005). Knowing
that there are at least two, you get more than double that probability (.0013),
but still not a very large value.
A: Any hand with 5 spades has 2 spades so it is not a really and intersection problem.
First take out 2 spades.  Now you have 50 cards left and 11 spades left.  You need 3 more spades.  
(11,3) / (50,3) = 0.84%
$\frac{11 \choose 3}{{50 \choose 3}} = 0.0084$
