Probability of transfers Probability of transferring materials between containers. 

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: $$P(X=1|G)=\frac{P(G|X=1)\cdot P(X=1)}{P(G)}=\frac{\frac{1}{3}\frac{28}{55}}{\frac{1}{3}}=\frac{28}{55}$$
