Finding the probability of selecting a certain ball on the $j$th draw from $M$ balls with $W$ white balls There are several parts to this question, but I feel that if I understand the first part I might be able to understand the other parts.
The question I am given is: In a box containing $M$ balls, of which $W$ are white, a sample of $n$ balls is drawn without replacement. Let $A_j$, where $j=1,2,...,n$ represent the event that the ball drawn on the $j$th draw is white. Let $B_k$ denote the event that the sample contains k white balls. Find the probability of $A_j$. 
This is what I've understood so far:
$$\begin{align} \text{selecting a white ball first: }P(A_1)&=\frac{W}{M}\\\text{selecting a non-white ball, then a white ball: }P(A_2)&=\frac{M-W}{M}+\frac{W}{M-1}\\\text{selecting two non-white, then a white ball: }P(A_3)&=\frac{M-W}{M}+\frac{M-W-1}{M-1}+\frac{W}{M-2}\end{align}$$ Is this train of thought correct? If it is, I'm not really seeing a way to deduce $P(A_j)$. Instead of this, should I be thinking about $\frac{M!}{(M-n)!}$ because order seems to be important and I'm selecting $n$ balls from the total $M$ balls?
I'm really confused at this point, and any help would be greatly appreciated. I tried to search other probability problems that were similar, but couldn't find anything. Thanks for any hints or links to similar problems in advance!

In order to determine the probability of an "and" event happening, you have to multiply the probabilities, not add them.
Try the hyper-geometric distribution if you have learned it
