<s>[INST] 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.
 [/INST] The question can be considered as two events' intersection, that a). $X$ white balls drawn from $U_1$ and $U_2$; b). the ball drawn from $U_3$ is white.
For the probability of event (a), we can consider it as a hypergeometric case. We’ve already known that there’re 3 white balls totally, so X∈(0,1,2,3). Applying the PMF of hypergeometric distribution:
$P(X)=\left(\frac {{3 \choose X}{4 \choose 4-X}}{{7\choose 4}}\right)$
In terms of event b, it’s nothing more than $ P(white)$=$X/4$
So what you need to do is:
$P=\sum_{X=0}^{4} P(X)P(white)$
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