Welp, I'm finding it a bit difficult to draw a tree diagram on here. :-) Are you able to set that up OK? That will really, really help with the visual on this so you can see where everything is.
But I'll try to do it without a tree and get you started. If you can get the tree done, put the probabilities already given on it.
Remember that all probabilities must add up to 1.00. So let's start with, if A and B are the only outcomes to the first stage (i.e. first set of branches on your tree), and we know what A is, then what does B have to be?
Now if you put your C and D branches off the A, the C branch is .48 as given in the problem. Keep in mind that "given" means that the probably of C happening is .48 ONLY IF we know A has already happened. That is, B doesn't count for that.
A (.3) --> .48 C
OR --> ? D
So that .48 is out of the .3 only, because we've eliminated it down to just A. So what is .48 of .3? That's the probability of C.
Then, keeping in mind that all probabilities add up to 1, what's the probability of D on the A branch only? (That's D | A).
You will also have a B branch and should be able to figure out the probability of B. Then put C and D off that as well. Insert the D given B from the problem, and kinda work the same way as A.
Just see if you can set up that tree to get the visual and make some sense of this, and see what happens. :-)