The probality of taking first blue and last red ball when picking 6 balls In the post the question was:
There is $36$ balls. $12$ are red R, $12$ are blue B, $12$ are yellow.
What is the probability of taking first blue and last red ball when picking $6$ balls? (Not returning them back).
There is a quick answer from André Nicolas.

Think of the balls as having distinct ID numbers. All sequences of $6$ balls (thinking of ID numbers, not colours) are equally likely.
The probability the first is blue is $\frac{12}{36}$. The probability the sixth is red, given the first was blue, is $\frac{12}{35}$. Multiply.

But I can't still visualize.
(1) If the position doesn't matter, then why the probability of the last ball is red is not equal to $\frac{12}{36}$?
(2) If given the first was blue matters, then why aren't we considering the middle balls so that the probability of the last ball is red is equal to $\frac{12}{31}$?
(3) doesn't The situation become complex if the middle ball(s) is/are red?
 A: (1) The probability the last ball is red is 12/36. Like the answerer emphasized, the probability the last ball is red given the first is blue is 12/35. The denominator is 35 cause there's only 35 balls left after picking the first one and the numerator is 12 since there's 12 red balls left (since the first ball chosen was not red).
(2) Why would we consider the middle balls? We have no information about them and thus do not need to condition on them at all.
(3) No. Of course if we learned of the information of what the middle balls were, it would affect the conditional probability of the last one being red, but we don't. We are "averaging over" the middle balls.
In order to draw the first ball blue and the last ball red, what needs to happen is:
a. The first ball we pick is blue (prob $\frac{12}{36}$)
b. and then, after picking the first ball blue, the last one needs to be red. You can think of this as a new experiment where theres 11 blue, 12 yellow and 12 red... what's the probability the last ball is red then? It's 12/35.
The desired outcome occurs iff both these things happen. Thus the probability is (12/36)*(12/35)
A: Instead of choosing balls randomly, imagine them laid out randomly in a row,
and choose #$1$ and #$6$. [It's the same thing ]
P(first ball is blue) $= \frac{12}{36}$ 
Now look at the situation where $35$ balls including $12$ red are left at serials $2-36$
Since balls don't have any preference for positions, a red ball now has a $Pr =\frac{12}{35}$ of occupying any of these positions, hence a $Pr=\frac{12}{35}$ of being at #6,
i.e. P(sixth red | first blue ) $=\frac{12}{35}$
You should be able now to see that P(first blue $\cap$ sixth red ) $= \frac{12}{36}\frac{12}{35}$ 
Furthermore, P( first yellow $\cap$ sixth red) $= \frac{12}{36}\frac{12}{35}$,
and P(first red $\cap$ sixth red) $= \frac{12}{36}\frac{11}{35}$  
