Compute the probability mass function of X There are two bags A and B. Bag A has 2 green marbles and 3 red marbles. Bag B has 2 green marbles and 3 blue marbles. Two marbles are drawn without replacement from bag A and put in bag B. Then two marbles are drawn without replacement from bag B and put in bag A. Let X be the number of green marbles in bag A after the above operations. Compute the probability mass function of X.
My attempt:
$X=\{0,1,2,3,4\}$
$X=0$: take all green from A and put non-greens back = $\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{7}\cdot\frac{2}{6}=\frac{17}{70}$
$X=1$: take all green from A and put one green back + take 1 green and put non-greens back = $\left(\frac{2}{5}\cdot\frac{1}{4}+\frac{4}{7}\cdot\frac{3}{6}\right)+\left(\frac{2}{5}\cdot\frac{3}{4}+\frac{4}{7}\cdot\frac{3}{7}\right)=\frac{228}{245}$
Then we continue on with $X=2,3,4$. My concern is that if we add up $\mathbb{P}(X=0)$ and $\mathbb{P}(X=1)$, we get a value that is greater than $1$, which should not be the case because everything should sum to $1$. Can someone tell me what I am doing wrong and possibly how I would approach this problem differently to get the right answer? Thank you.

You wrote:

take all green from A and put non-greens back

First sight first error found...
$$\mathbb{P}[X=0]=\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{70}$$
At present I did not check the rest
