There are $2$ boxes full of black and white balls. The first one has $a$ white and $b$ black balls ($a\geq2$ && $b\geq2$). The second one has $c$ white and $d$ black balls ($c\geq1$ && $d\geq1$). You draw $2$ balls from the first box and $1$ ball from the second box and put them into a third box. What is the probability to draw a white ball from the third box.
I tried to solve it this way:
$A_1$ - draw 2 white balls from first box and 1 white from second box and put them in third box.
$A_2$ - draw 2 white balls from first box and 1 black from second box and put them in third box.
$A_3$ - draw 1 white ball and 1 black ball from first box and 1 white from second box and put them in third box.
$A_4$ - draw 1 white ball and 1 black ball from first box and 1 black from second box and put them in third box.
$A_5$ - draw 2 black balls from first box and 1 white from second box and put them in third box.
$A_6$ - draw 2 black balls from first box and 1 black from second box and put them in third box.
$B$ - draw white ball from third box.
$$P(B) = \sum_{i = 1}^{6} P(B \mid A_i)P(A_i)$$
I think $P(B\mid A_i)$ are pretty obvious so i won't write them.
After long brute force i came up with this answer
$$\frac{ac(5(a-1) + 6b + b(b-1))}{3(a+b)(a+b-1)(c+d)}$$
I want to know if there is a shorter answer.
I'm not sure where your brute force went wrong, but the answer is much simpler:
$${2\over3}\cdot{a\over a+b} + {1\over3}\cdot{c\over c+d}$$
That is, the chosen ball came from the first box with probability $2/3$ and from the second box with probability $1/3$. More generally, if you fill the third box with $m$ balls from the first box (with $0\le m\le a+b$) and $n$ ball from the second box (with $0\le n\le c+d$), the probability of drawing a white ball from the third box is
$${m\over m+n}\cdot{a\over a+b}+ {n\over m+n}\cdot{c\over c+d}$$