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.