Three boxes with balls Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that the ball is white.

Well, I reasoned in this way. The possible combinations that ensure that $U_3$ contains at least one white ball are BNBB, NBBB, NNBB, BNBN, BNNB, BNNN, NBBN, NBNB, NBNN, NNBN, NNNB. Thus:

*

*$\mathbb{P}$($U_3$ contains $3$ white balls)$=\mathbb{P}($(BNBB)$\cap$(NBBB)$)=(\frac{1}{3}\cdot1 \cdot\frac{1}{2}\cdot\frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot \frac{1}{3})=0,11$


*$\mathbb{P}(U_3$ contains $2$ white balls)$=\mathbb{P}($(NNBB)$\cap$(BNBN)$\cap$(BNNB)$\cap$(NBBN)$\cap$(NBNB)$)=(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{3}\cdot 1\cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{1}{3}\cdot 1 \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{2}{3})=0,46$


*$\mathbb{P}(U_3$ contains $1$ white ball)$=\mathbb{P}($(BNNN)$\cap$(NBNN)$\cap$(NNBN)$\cap$(NNNB)$)=(\frac{1}{3}\cdot 1 \cdot \frac{1}{2}\cdot \frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{2}{3})=0,33$


*$\mathbb{P}(U_3$ doesn't contain any white balls)$=2\mathbb{P}($(NNNN)$)=2(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})=0,1$
Thus $\mathbb{P}($one white ball from $U_3)=\frac{3}{4}\cdot 0,11+\frac{2}{4}\cdot 0,46+\frac{1}{4}\cdot 0,33+\frac{0}{4}\cdot 0,11=0,395$
Is it correct? Particularly I'm interested in reasoning. Thanks in advance.

Let $X_1, X_2$ be the count of white balls drawn from the first two urn, $Y$ be the sum of these, and $E$ the event that a white ball is then drawn from the third urn.
Your approach was correct.  $\mathsf P(E)=\sum_{k=0}^4 \mathsf P(E\mid Y=k)\mathsf P(Y=k)$
However, your evaluations were a bit off.
$$\begin{align}\mathsf P(Y=3)&=\mathsf P(X_1=1)\mathsf P(X_2=2)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 22\binom 20}{\binom 42}\\&=\dfrac{1}{9}\\[2ex]\mathsf P(Y=2)&=\mathsf P(X_1=1)\mathsf P(X_2=1)+\mathsf P(X_1=0)\mathsf P(X_2=2)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 21\binom 21}{\binom 42}+\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 22\binom 20}{\binom 42}\\&=\dfrac 12\\[2ex]\mathsf P(Y=1)&=\mathsf P(X_1=1)\mathsf P(X_2=0)+\mathsf P(X_1=0)\mathsf P(X_2=1)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 20\binom 22}{\binom 42}+\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 21\binom 21}{\binom 42}\\&=\dfrac 13\\[2ex]\mathsf P(Y=0)&=\mathsf P(X_1=0)\mathsf P(X_2=0)\\&=\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 20\binom 22}{\binom 42}\\&=\dfrac 1{18}\end{align}$$
