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...
 A: 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.)
A: You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.
If $N$ is the number of dictionaries where the keyword is found then $$\Pr(N=n) = \frac{{5 \choose n}{4 \choose 4-n}}{9 \choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall. 
So for $n=0,1,2,3,4$ this gives probabilities $\frac{1}{126},\frac{20}{126},\frac{60}{126},\frac{40}{126},\frac{5}{126}$.
Add up the last three (and change to lowest terms) and you have your solution.
A: Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword. 
Let $c=4$ databases are chosen and $s$ databases have the keyword.
The question is asking to find $s\ge 2$, i.e. $s=\color{red}2,\color{green}3,\color{blue}4$.
There are ${t\choose c}={9\choose 4}$ ways to choose $c$ databases from total $t$ databases.
Case 1: There are ${k\choose 2}={5\choose \color{red}2}$ ways to choose from $k$ and ${n\choose 2}={4\choose 2}$ ways from $n$, hence: ${5\choose 2}{4\choose 2}$.
Case 2: There are ${k\choose 3}={5\choose \color{green}3}$ ways to choose from $k$ and ${n\choose 1}={4\choose 1}$ ways from $n$, hence: ${5\choose 3}{4\choose 1}$.
Case 3: There are ${k\choose 4}={5\choose \color{blue}4}$ ways to choose from $k$ and ${n\choose 0}={4\choose 0}$ ways from $n$, hence: ${5\choose 4}{4\choose 0}$.
Hence, the required probability is:
$$P(s\ge 2)=P(s=2)+P(s=3)+P(s=4)=\\
\frac{{5\choose 2}{4\choose 2}}{9\choose 4}+\frac{{5\choose 3}{4\choose 1}}{9\choose 4}+\frac{{5\choose 4}{4\choose 0}}{9\choose 4}=\\
\frac{60+40+5}{126}=\frac{35}{42}\approx 0.83.$$ 
Note that you can also use the complement, i.e.:
$$P(s\ge 2)=1-P(s=0)-P(s=1).$$
Thus, in general, the formula is:
$$P(s\ge r)=\sum_{i=r}^c P(s=i)=\sum_{i=r}^c \frac{{k\choose i}{n\choose c-i}}{{t\choose c}}.$$
Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.
