Probability assignment There are two boxes. In the first one there are, 4 black and 3 white balls, and in the second one there are 3 black and 5 white balls. We take two balls from the first box, and one from the second one and we put then in a third box. If we take out one ball from the third box, what is the probability that the ball we took out is white?
Can someone help me with this?

Let $A$ be the event that a white ball is taken from box 3. Also let $X_k$ be the event that $k$ white balls are in box 3. 
Using the law of total probability we get, 
$P(A) = \sum_{k=0}^3P(A|X_k)P(X_k)$.
$P(A|X_k)$ are easy to calculate. If there are NO white balls in box 3, that is $k=0$, then $P(A|X_0) = 0$. If there are 1 white balls in box 3 ($k=1$), then $P(A|X_1)=1/3$. Similar argument shows that $P(A|X_2) = 2/3$ and $P(A|X_3) = 1$.
Next we calculate $P(X_k)$.
$P(X_3)$ = $P$(2 white balls from box 1 AND 1 white ball from box 2). Since taking ball(s) from box 1 and box 2 are independent events we have
$P$(2 white balls from box 1 AND 1 white ball from box 2) = $P$(2 white balls from box 1)$P$(1 white ball from box 2)
$P$(2 white balls from box 1) =  $\frac{3\choose2}{7\choose2}$
$P$(1 white ball from box 2) = $5/8$.
Hence $P(X_3) = \frac{3\choose2}{7\choose2}\frac{5}{8} = \frac{5}{56}$
Similar arguement shows that 
$P(X_2) = \frac{3\choose2}{7\choose2}\frac{3}{8} + \frac{{3\choose1}{4\choose1}}{7\choose2}\frac{5}{8} = \frac{23}{56}$.
$P(X_1)= \frac{4\choose2}{7\choose2}\frac{5}{8} + \frac{{3\choose1}{4\choose1}}{7\choose2}\frac{3}{8} = \frac{22}{56}$.
We do not have to calculate $P(X_0)$ since it will be cancelled out by $P(A|X_0)$ which equals $0$.
Finally, putting it all together,
$P(A) = 0 + \frac{1}{3}\frac{22}{56} + \frac{2}{3}\frac{23}{56} + \frac{5}{56}$
which is approximately $0.49404$       
