Conditional probability problem An Internet search engine looks for a keyword in 9 databases, searching them in random order. Only 5 of these databases contain the given keyword. What is the probability that it will be found in at least 2 of the first 4 searched databases?
What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases,
then we need to find a conditional probability 
$P(X\geq 2 | Y)=P(X= 2,3,\text{ or }4 | Y)=\frac{P(\{X\geq 2\} \cap Y)}{P(Y)}$
 but then I don't know how to continue?
Any help...

Added: Henry's and my answers, while the approaches are different, actually give the same result!  To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases.  Let $X$ be the number of the $K$ databases chosen that contain the keyword.  For $P(X = x)$, Henry's and my logic give the following answers.
$$\text{ Henry: } P(X = x) = \frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}}; \text{    me: } P(X = x) = \frac{\binom{K}{x} \binom{N-K}{M-x}}{\binom{N}{M}}.$$
But
$$\frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}} = \frac{M!}{x! (M-x)!} \frac{(N-M)!}{(K-x)! (N-M-K+x)!} \frac{K! (N-K)!}{N!}$$
$$= \frac{K!}{x! (K-x)!} \frac{(N-K)!}{(M-x)! (N-K-M+x)!} \frac{M! (N-M)!}{N!} = \frac{\binom{K}{x} \binom{N-K}{M-x}}{\binom{N}{M}}.$$
Thus the two answers give the same result.

(Original answer.)  
I would approach it differently.  Let $X$ be the number of the first four databases in which the keyword is found.  You want $P(X \geq 2)$.  
$$P(X = 2) = \frac{\binom{4}{2} \binom{5}{3}}{\binom{9}{5}}$$ because there are $\binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $\binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $\binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.
I'll let you calculate $P(X = 3)$ and $P(X = 4)$.
(FYI, $X$ has what's called a hypergeometric distribution.)
