probability: Urn question with replacement and additional balls sorry about this as i know ive been posting quite a bit recently but im hoping that probability may have "clicked" so i just want to make sure i'm getting this right, as i know of few places better than Maths stack i thought i'd pick your brains once again.
From a first course in probability: Page 102: Question 3.14
An Urn initially contains 5 white balls and 7 black balls. Each time a ball is selected its colour is noted and it is replaced in the urn along with 2 other balls of the same colour. What is the probability that:
A) the first two balls selected are black and then the next two are white.
B) of the first 4 balls selected, exactly 2 are black.
Solution:
Let A be the event that the first ball drawn is black.
Let B be the event that the second ball drawn is black
Let C be the event that the third ball drawn is white
Let D be the event that the fourth ball drawn is white.
then $P(A \cap B \cap C \cap D)=P(A)P(A|B)P(C|A\cap B)P(D|A \cap B \cap C)$
so we have
$P(A)=\frac{7}{12}$ 
Let B be the event that the second ball drawn is black.$P(B|A)=\frac{P(B \cap A)}{P(A)}$
the event $B \cap A$ is then the event in which a black ball is drawn and then a second black ball is drawn from the new urns composition. as such i make this
$P(B \cap A) = \frac{{{7}\choose{1}}{{9}\choose{1}}}{{{12}\choose{1}}
{{14}\choose{1}}}$ which is the number of ways of picking 1 black ball from 7 then 1 black ball from 9 and this so happens to be $P(A)P(B)$ suggesting that we are taking these balls independantly.
in which case
$$P(B|A)=P(B),~ P(C|A\cap B)=P(C), ~ P(D|A \cap B \cap C)=P(D)$$
this leads me to.
$$P(A \cap B \cap C \cap D)=P(A)P(A|B)P(C|A\cap B)P(D|A \cap B \cap C)=P(A)P(B)P(C)P(D)=\frac{7}{12}\cdot\frac{9}{14}\cdot\frac{5}{16}\cdot\frac{7}{18} =\frac{35}{768}$$
which is the right answer. but getting the right answer via the wrong method is no good for me so i want to make sure this has been calculated soundly.
For Part B. i reasoned that so long as we draw exactly 2 balls which are black and exactly 2 which are white then the probability of A wont change reguardless of what order we draw them. in which case we simply need to add up the different ways of drawing these four balls. giving me
$$\frac{4!}{2!\cdot 2!}\cdot \frac{35}{768} = \frac{210}{768}$$
which again is the right answer.
anyone willing to go through this would be greatly appreciated by me. and as such thank you in advance.
 A: The events are not independent.   You have reasoned that $\mathsf P(B\mid A)=9/12$ and likewise should reason that $\mathsf P(B\mid A^\complement)=5/12$.
Thus by the Law of Total Probability: $\mathsf P(B)~{=\mathsf P(A)\,\mathsf P(B\mid A)+\mathsf P(A^\complement)\,\mathsf P(B\mid A^\complement) \\ = \tfrac 7{12}\tfrac 9{14}+\tfrac {5}{12}\tfrac{7}{14} \\ = \tfrac 7{12} }$
Thus, $\mathsf P(A\cap B)\neq \mathsf P(A)\,\mathsf P(B)$ .
Fortuitously though, you have actually used the correct evaluation for part (a).
$$\mathsf P(A\cap B\cap C^\complement\cap D^\complement)~{=\mathsf P(A)\,\mathsf P(B\mid A)\,\mathsf P(C^\complement\mid A\cap B)\,\mathsf P(D^\complement\mid A\cap B\cap C^\complement) \\ = \tfrac 7 {12}\tfrac 9{14}\tfrac 5{16}\tfrac 7{18} }$$
Note: here I use that $A,B,C,D$ are the events for drawing black on the first to fourth draw, respectively, so $C^\complement,D^\complement$ are the events for drawing white on the third and forth draws.

For part (b), the probability for drawing some arrangement of two black and two white will be $${\quad{\mathsf P(A{\cap}B{\cap}C^\complement{\cap}D^\complement)+\mathsf P(A{\cap}B^\complement{\cap}C{\cap}D^\complement)+\mathsf P(A{\cap} B^\complement{\cap}C^\complement{\cap}D)+\mathsf P(A^\complement{\cap}B{\cap} C{\cap}D^\complement)+\mathsf P(A^\complement{\cap}B{\cap}C^\complement{\cap}D)+\mathsf P(A^\complement{\cap}B^\complement{\cap}C{\cap}D)}
\\={\dfrac{7\cdot 9\cdot 5\cdot 7+7\cdot 5\cdot 9\cdot 7+7\cdot 5\cdot 7\cdot 9+5\cdot 7\cdot 9\cdot 7+5\cdot 7\cdot 7\cdot 9+5\cdot 7\cdot 7\cdot 9}{12\cdot 14\cdot 16\cdot 18}} \\ ={\dfrac{4!}{2!\cdot 2!}\cdot\dfrac{5\cdot 7^2\cdot 9}{12\cdot 14\cdot 16\cdot 18}}}$$
Or, in short: for each of the six orders in which you may draw the four balls occurs, the count is incremented, once for each colour, by two.   As you correctly reasoned, thusly the probability for each arrangement is identical, and there are six of them.
