Probability of transfers Probability of transferring materials between containers. 
 A: Define the random variable: $X=$ is the number of green marbles drawn, which has the hypergeometric distribution with parameters $N=8+4=12$, $K=4$ and $n=3$. Then you have that the probability mass function of $X$ is equal to:  $$P(X=k)=\frac{\dbinom{4}{k}\dbinom{8}{3-k}}{\dbinom{12}{3}}$$ for $k=0,1,2,3$. So, by substitung in the above formula we find that: $$P(X=0)=\frac{14}{55}, P(X=1)=\frac{28}{55}, P(X=2)=\frac{12}{55}P(X=0)=\frac{1}{55}$$
1st Question. Define the event $G$ the marble drawn from the second jar. According to the Total Probability Law you have (by conditioning on X) that:
$$\begin{align*}P(G)&=P(G|X=0)P(X=0)+P(G|X=1)P(X=1)+P(G|X=2)P(X=2)\\&\quad+P(G|X=3)P(X=3)=\\&=0P(X=0)
+\frac{1}{3}P(X=1)+\frac{2}{3}P(X=2)+1P(X=3)=0+\frac{1}{3}\frac{28}{55}+\frac{2}{3}\frac{12}{55}+\frac{1}{55}\\&=\frac{1}{3}\end{align*}$$ 
2nd Question. You want to calculate the probability: $P(X=1|G)$. According to Bayes theorem you have that: $$P(X=1|G)=\frac{P(G|X=1)\cdot P(X=1)}{P(G)}=\frac{\frac{1}{3}\frac{28}{55}}{\frac{1}{3}}=\frac{28}{55}$$
A: The situation sketched in the question is equivalent to the one sketched here:

$3$ marbles are drawn from a jar that contains $8$ white and $4$ green marbles and the color of the marble that is drawn firstly is checked. 

Questions are: 
1) What is the probability that the checked color is green? Immediate answer: $\frac{4}{12}=\frac{1}{3}$.
2) If the checked color is indeed green then what is the probability that the color of the other drawn marbles is white?
If the checked color is green then after the first draw $11$ marbles were left in the jar from which $8$ are white. So the probability for the marble drawn secondly to be white is $\frac{8}{11}$. Under the assumption that the second has indeed color white  the probability that the third is white also is $\frac{7}{10}$. So the probability that both are white is $\frac{8}{11}\times\frac{7}{10}=\frac{28}{55}$. This answers 2).
