How to decide whether to use Bayes' theorem or not? 
A bag contains 12 red balls  and 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in next two drawn , exactly one white ball is drawn. ( Leave answer in $ \binom{n}{r}$)

What I did:
$$ P = \frac{\binom{6}{4} \binom{12}{2}}{\binom{18}{6}}  \frac{\binom{2}{1} \binom{10}{1}}{\binom{12}{2}}+  \frac{\binom{6}{5} \binom{12}{1}}{\binom{18} {6}} \frac{\binom{1}{1} \binom{11}{1}}{\binom{12}{2}} +  \frac{\binom{6}{6} \binom{12}{0}}{\binom{18}{6}} \cdot 0$$
The idea behind this: so the first factor in each term of sum, is the probability for a number of white ball- red ball combination to be drawn and the factor multiplied with it is the probability for drawing one white ball from the remaining pool.
However the solution key says,
$$ P= \frac{\binom{12}{2}\binom{6}{4} \binom{10}{1} \binom{2}{1} + \binom{12}{1} \binom{6}{5} \binom{11}{1}\binom{1}{1} }{\binom{12}{2} ( \binom{12}{2} \binom{6}{4} + \binom{12}{1} \binom{6}{5} + \binom{12}{0} \binom{6}{6})}$$
This seems to be some sort of Bayes' theorem application but I can't really see why we would need Bayseian thinking here.
 A: First a formula: Suppose we have $R$ red balls & $W$ white balls, then the probability of selecting a specific sequence of $r $ red balls & $w $ white balls (with $r \le R, w \le W$) is
$ { (R+W - (r+w))! \over (R+W)!} { R! \over (R-r)! } {W! \over (W-w)! }$. Since these can
be rearranged in $\binom{w+r}{w}$ ways, we see that the probability of selecting $r $ red balls & $w $ white balls (in any order) is
$p((R,W),(r,w)) = \binom{w+r}{w} { (R+W - (r+w))! \over (R+W)!} { R! \over (R-r)! } {W! \over (W-w)! } = {1 \over \binom{W+R}{w+r} }\binom{W}{w} \binom{R}{r}$.
Let $B$ be the event that the $7$th and $8$th draw have exactly one white ball.
Let $A_k$ be the event that exactly $k$ white balls are drawn in the first $6$ draws.
Let $A= A_4 \cup A_5 \cup A_6$ (disjoint union).
We want to compute $P[B|A] = {P[B \cap A] \over P[A] } = {P[B \cap A_4]+P[B \cap A_5]+p[B \cap A_6] \over P[A_4]+P[A_5]+P[A_6]} = {P[B|A_4]P[A_4]+P[B|A_5]P[A_5]+p[B|A_6]P[A_6] \over P[A_4]+P[A_5]+P[A_6]} $.
Note that $P[B|A_k] = p((10+(k-4),2-(k-4)),(1,1))$.
Substituting we get
$P[B|A]= { p((10,2),(1,1)) p((12,6),(2,4))+p((11,1),(1,1)) p((12,6),(1,5))+p((12,0),(1,1)) p((12,6),(0,6)) \over p((12,6),(2,4))+p((12,6),(1,5))+p((12,6),(0,6))} $.
Substituting for $p$ gives the solution key answer above.
