Trivial Probability?

[StoneBlack]

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A. 1/4
B. 1/2
C. 5/8
D. 2/3
E. 3/4

This question is from 300+ GMAT questions document available on BeatTheGMAT forums.
The solution is provided there but I am not really convinced of it.

[shankar.ashwin]

Each ball picked has a 3/8th probability of being yellow and 5/8th probability of being black. The number of selection doesn't matter.

Ofcourse theres a mathematical proof which is lengthy, but think about it, each ball is likely to be either yellow or black, now how does the number of pick matter..

If the probability was any greater or lesser you would have to say, the first 3 picks were biased towards picking a particular color, which is not the case. Hence probability at any pick shouldn't change.

P (nth pick) = P (1st pick) = [spoiler]5/8[/spoiler]

[GMATGuruNY]

Check my posts here:

https://www.beatthegmat.com/probablity-ques-t60161.html

[StoneBlack]

Thanks for the pointer.
Apparently i tried the expanded way by considering all possibilities:
3Y,1B
2Y2B
1Y3B
4B

but failed to take into consideration the different arrangement possible in 2Y2B and 1Y3B.
Thanks again!