A bag contains 5 white and 7 black balls . 3 balls are drawn at random . Find the probability that (I) all are white (ii) one white and 2 black

Are they being drawn at the same time. As there are no instructions on that, I'll assume they are not.
I don't know the name of what I'll be using in english but you have got to do a diagram.
On the 1st part of the diagram, it is divided in 2 parts (one for white and one for black). The white starts with 5/12 (probability of the 1st ball being white), then it is 4/11, assuming the 1st ball was white, then ends with 3/10. Now 5/12*4/11*3/10=60/1320 which is equal to 6/132=3/66=1/12. This is the probabilty of all of them being white.
Now for the second part of the question, the first ball is white, therefore it starts with 5/12. But now the second one is black, so it is 7/11 (the 7 black balls are still there, but there are 11 in total because the white was removed). The third one will therefore be 6/10. Once again, 5/12*7/11*6/10= 210/1320= 21/132= 7/41.
I don't renember this quite well because it was sometime ago, but I think this is right.