text: Q:
  
  Dice, balls and boxes probability problem (conditional probability)
  
  Problem
  Suppose there are two boxes $A$ and $B$ such that $A=\{\text{5 red balls and 3 white balls}\}$, $B=\{\text{1 red ball and 2 white balls}\}.$
  A dice is thrown, if the result is $3$ or $6$, a ball from the box $A$ is extracted and then placed it in box $B$ and after that a ball from box $B$ is selected. If the result of the dice is neither $3$ nor $6$, then the inverse procedure takes place (a ball from box $B$ is chosen, then is is thrown in box $A$ and afterwards a ball from box $A$ is selected).
  Calculate the probability of both balls being red.
  The attempt at a solution
  I've denoted $R=\{\text{both balls extracted in each step are red}\}.$, $D_{3,6}=\{\text{the result of the dice is 3 or 6}\}.$ I can calculate $P(R)=P(R|D_{3,6})+P(R|{D_{3,6}}^c).$
  Using the formula for conditional probability, I have that $$P(R|D_{3,6})=\frac{P(R \cap D_{3,6})}{P(D_{3,6})},$$ $$P(R|{D_{3,6}}^c)=\frac{P(R \cap {D_{3,6}}^c)}{P({D_{3,6})}^c}.$$
  I've calculated $P(R \cap D_{3,6})=\frac{5}{8}\frac{2}{4}$, and $P(D_{3,6})=\frac{2}{6}$.
  And $P(R \cap {D_{3,6})}^c)=\frac{1}{3}\frac{6}{9}$, $P({D_{3,6})}^c)=\frac{4}{6}$.
  Using these results, we obtain $$P(R)=\dfrac{\frac{5}{8}\frac{2}{4}}{\frac{2}{6}}+\dfrac{\frac{1}{3}\frac{6}{9}}{\frac{4}{6}}.$$
  In order to know if my result at least makes sense, I've calculated the actual value of that and it is more than $1$, so there is something wrong with my solution. I am looking for a corrected answer and/or an alternative solution. Thanks in advance.
  
  A:
  
  You can use
  $P(R)=P(R|D_{3,6})P(D_{3,6})+P(R|D_{3,6}^{c})P(D_{3,6}^{c})=\left(\frac{5}{8}\cdot\frac{1}{2}\right)\cdot\frac{1}{3}+\left(\frac{1}{3}\cdot\frac{2}{3}\right)\cdot\frac{2}{3}=\frac{109}{432}$.
  
  