I'll assume that it is not going to search the same database twice.
The denominator will be C(9,4) = 126

There are 4 that don't contain the password and 5 that do.

We find the probability of the complement event, that the first 4 searches
do not find at least 2 that contain the password at least twice.  That means
that either it finds none that contain it, or it finds exactly 1 that contains
it.

The number when exactly 0 contain the password is 1 way. Choose all 4 that don't
contain the password.  C(4,4)=1

The number when exactly 1 contains the password.
For each of the C(4,3)=4 choices of 3 that don't contain the password, 
we choose 1 that does in C(5,1)=5 ways.  That's (4)(5)=20 ways.

That's 1+20 = 21 unacceptable ways.

So the probability of the complement event is 21/126 = 1/6

So the probability of the desired event is 1-1/6 or 5/6.

Edwin

For me, it is easier to reformulate the problem.

I will consider the databases as balls marked by letters G (=good) if they contain the keyword,
or B (=bad) if they do not contain.

So, I have 5 balls marked G and 4 balls marked B.


    +--------------------------------------------------------------------------+
    |    For clarity, I consider all 9 balls as different (distinguishable),   |
    |        but the order of balls in my selections does not matter,          |
    |       as in classic combinations. Only their amounts do matter.          |
    +--------------------------------------------------------------------------+


I will consider the first 4 searched databases as box N (= oNe) and the rest 5 databases
as box T (= Two).


So, now I have 5 balls G and 4 balls B.
I also have box N (meaning first 4 searched databases) and box T (meaning the rest 5 databases).


        An experiment is placing arbitrary (= randomly) 4 of 9 balls into box N 
            (and placing the rest 5 balls in box T just without choice).

    The question is to find the probability of getting 2, or 3, or 4 balls G in box N.


Total of possible different combinations in box N is   = 126.

It is the size of the space of events in my consideration.


The number of combinations getting 2 balls G in box N is   = 10*6 = 60

                           (we combine 2G of 5G and 2B of 4B in box N)


The number of combinations getting 3 balls G in box N is   = 10*4 = 40

                           (we combine 3G of 5G and 1B of 4B in box N)


The number of combinations getting 4 balls G in box N is   = 5*1 = 5

                           (we combine 4G of 5G and 0B of 4B in box N)


Now the answer to the problem's question is

    P =  =  =  = .


ANSWER.  The probability is  .