Picking balls at random until getting a special one I could really use some help in understanding this homework problem.
I am sorry in advance for English mistakes, I'm not a native English speaker :)
The question:
There are 20 balls. 3 of them are white and the rest are blue. What is the probability that a white ball is picked for the first time in $k^{th}$ trial?
My attempt at solution:
I tried to calculate the total number of possibilities for choosing $k$ balls out of 20 when 3 are different from the rest but I failed to do that (an explanation could help me here).
Instead, I tried looking at a (hopefully?) equivalent problem: arranging the balls in a row. I calculated the number of possible different arrangements of all the 20 balls in a row:
$$\binom{20}{3}$$
Now I don't understand how to calculate the number of arrangements out of them that are equivalent to picking $k$ balls until getting a white one.
I think that I should consider all the possible arrangement so that the $k^{th}$ ball is white. There are $\binom{20-k}{2}$ such arrangements but I am not sure. Does the arrangement of balls after the $k^{th}$ matter and should be considered?
Any help would be greatly appriciated!
 A: You seek the probability that the $k$th ball is the first white ball.  

Instead, I tried looking at a (hopefully?) equivalent problem: arranging the balls in a row. I calculated the number of possible different arrangements of all the 20 balls in a row:

Yes. Excellent.
So you seek the probability that when placing the three white balls among $20$ drawing positions, they will be placed: none among the first $k-1$ places, one in the $k$-th place, and two among the latter $20-k$.
Indeed this is: $$\begin{align}\dfrac{\dbinom{k-1}{0}\dbinom 11\dbinom{20-k}2}{\dbinom {20}3}&=\dfrac{\dbinom {20-k}2}{\dbinom{20}3}\end{align}$$

Does the arrangement of balls after the kth matter and should be considered?

You considered them when weighting the total outcome set, so you should also consider them when weighting the favoured set, because they do affect the weighting of the outcomes (by considering them you ensure all outcomes are equally probable).

Alternatively, you can evalute the probability of obtaining $k-1$ from $17$ blue balls, then $1$ from $3$ white balls, when drawing $k-1$ from all $20$ balls then $1$ from the remaining $21-k$ balls. 
$$\dfrac{\dbinom{17}{k-1}\dbinom{3}1}{\dbinom{20}{k-1}\dbinom{21-k}{1}}$$
This does not consider what happens in the latter draws, but...
$$\dfrac{\dfrac{17!}{(k-1)!(18-k)!}\dfrac{3!}{2!1!}}{\dfrac{20!}{(k-1)!(21-k)!}\dfrac{(21-k)!}{1!(20-k)!}}=\dfrac{\dfrac{(20-k)!}{2!(18-k)!}}{\dfrac{20!}{3!17!}}$$
