Find probability the 4th ball is white, then find probability all three were black The first part of this question is the same as this one.

A box has 10 balls, 6 of which are black and 4 of which are white. 3 balls are removed from the box, their color unnoted. Find the probability that a fourth ball removed from the box is white, Assume that the 10 balls are equally likely to be drawn from the box.

The answer is $\frac{2}{5}$ given we assume replacement.  However, I was looking at the question and the user stated they found their answer using binomial coefficients to show:
$\displaystyle\Pr(4^{th}\text{Ball is white})=\frac{\binom{6}{3}\frac47+\binom{6}{2}\binom{4}{1}\frac37+\binom{6}{1}\binom{4}{2}\frac27+\binom{4}{3}\frac17}{\binom{10}{3}}=\frac25$
Could someone explain how to use binomial coefficients and what is going on in the above equation?
Secondly, a follow-up question is:

Find the probability that all three of the removed balls will be black if it is known that at least one of the removed balls is black.

The answer is $\frac{5}{29}$, which I tried to solve for thinking about it as a $$Pr(B_1=Black)*Pr(B_2=Black|B_1 or B_2 = Black)*Pr(B_3=Black|B_1 or B_2 or B_3= Black)$$ but I could not solve for it because I got lost with the conditional probabilities and unions. I also tried to assume independence between each ball as $$\frac59*\frac59*\frac59$$ but was wrong as well. How do we solve for this?
 A: $\displaystyle\Pr(4^{th}\text{Ball is white})=\frac{\binom{6}{3}\frac47+\binom{6}{2}\binom{4}{1}\frac37+\binom{6}{1}\binom{4}{2}\frac27+\binom{4}{3}\frac17}{\binom{10}{3}}=\frac25$
$\binom{6}{3}\frac47$ is the number of ways to choose 3 black balls times the probability that the 4th ball is white given the first three are black
$\binom{6}{2}\binom{4}{1}\frac37$ is the number of ways to choose 2 black balls and 1 white ball times the probability that the 4th ball is white given the first three balls are (BBW)
$\binom{6}{1}\binom{4}{2}\frac27$ is the number of ways to choose 1 black balls and 2 white ball times the probability that the 4th ball is white given the first three balls are (BWW)
And finally, we choose 4 white balls.
This is divided by the number of ways to choose 3 balls from the set.
Now, I would say, that the probability that the n'th ball is white is the same as the probability that the 1st ball is white, and leave it at that.
A: Doug M (in the answers) and saulspatz (in the comments) have explained the first question well.
Sequentially removing the balls from the bag corresponds to placing the balls in sequence.  Since a white ball is equally likely to be in any particular position in the sequence, the probability that a white ball is drawn fourth is $4/10 = 2/5$ since four of the ten available balls are white.  

If three balls are removed from the bag, find the probability that all three of the removed balls are black given that at least one of the removed balls is black.

Our sample space consists of all selections of three balls in which at least one of the balls is black.  This consists of all selections of three balls except those in which none of the removed balls are black.  There are $\binom{10}{3}$ ways to select three of the ten balls.  If none of the three balls are black, then the three removed balls must be white.  There are $\binom{4}{3}$ ways to select three of the four white balls.  Thus, the number of selections of three balls in which at least one of the removed balls is black is 
$$\binom{10}{3} - \binom{4}{3}$$
The number of selections in which all three balls are black is
$$\binom{6}{3}$$
since we must select three of the six black balls.
Therefore, the conditional probability that all three balls are black given that at least one is black is 
$$\frac{\dbinom{6}{3}}{\dbinom{10}{3} - \dbinom{4}{3}}$$
