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.)