The other tutor's answer is wrong because he assumed something 
that was not given.  Notice that he wrongly assumed this:

We CANNOT assume this, for that would only be true if
A and B were given to be mutually exclusive.  But they
are only given to be dependent, not mutually exclusive!
If they were mutually exclusive then P(A|B)=0, not 0.55.
So he was definitely wrong to assume this. 

Since we are not given anything about sets AᑎB' or A'ᑎB', we cannot
find P(AᑌB).  However we can find P(AᑎB).  Wasn't that really what
you were asked to find, and there is a typo, and ᑌ should have been ᑎ?
If not, let me know in the thank-you note form below, and I'll get back
to you.

     P(A|B) = P(AᑎB)/P(B)

       0.55 = P(AᑎB)/0.4

(0.55)(0.4) = P(AᑎB)

       0.22 = P(AᑎB)

That is the answer if you were asked for P(AᑎB), not P(AᑌB)

-----------------------------------------------------------------
  
FYI, to show that to get P(AᑌB) is impossible, let's look at the 
Venn diagram below:



Let x,y,z and w be the probabilities of the 4 regions of the Venn
diagram.

P(B)= y+z = 0.4 

P(A|B) = P(AnB)/P(B) = y/(y+z) = y/0.4 = 0.55

 y/0.4 = 0.55
     y = (0.55)(0.4)
     y = 0.22 = P(AᑎB)

   y+z = 0.4
0.22+z = 0.4
     z = 0.18 = P(A'ᑎB)

We know that x+y+z+w = 1
             x+0.22+0.18+w = 1
             x+w = 0.6



So one possibility for x+w=0.6 is x=0.3 and w=0.3,

we could have:

 

Then P(AᑌB) = 0.3+0.22+0.18 = 0.7

But another possibility for x+w=0.6 is x=0.4 and w=0.2,

we could have:

 

Then P(AᑌB) = 0.4+0.22+0.18 = 0.8

And there are also infinitely more possibilities.

Let me know in the thank you note below if you still think you were 
asked to find P(AᑌB) and not P(AᑎB)
                                     
Edwin