The probability of 1 green marble and no red marble being drawn The question is

A jar contains 3 red, 2 white, and 1 green marble. Two marbles are drawn at random without replacement from the jar. Let $X$ represent the number of red marbles drawn and let $Y$ represent the number of green marbles drawn. What is $P(X=0,Y=1)$?

My idea is that we can either find the probability of pick a green marble on the first try and not a read marble on the second try. So the probability in this case would be 
\begin{equation}
\dfrac{1}{6} \times \dfrac{2}{5} = \dfrac{1}{15}
\end{equation} 
And the second case is not picking green or red on the first try and then picking green on the second try.
\begin{equation}
\dfrac{2}{6} \times \dfrac{1}{5} = \dfrac{1}{15}
\end{equation} 
Then adding them and we have 
\begin{equation}
P(X = 0, Y = 1) = \dfrac{1}{15} + \dfrac{1}{15} = \dfrac{2}{15}
\end{equation}
I am not sure if my approach is correct since in both two cases I am getting the same probabilities I feel like I am doing a step that is unnecessary. 
 A: Your approach is correct, since you are covering all possible events that satisfy the condition, and you are not over or undercounting.
An alternate perspective, which can confirm your answer, is that we have $ 6 \times  5=30 $ possible ways of conducting this draw. Of this, to satisfy our condition, we need the first ball to be white and the second one to be green, or vice versa. There are $ 2 \times 1$ ways to do both of the above, so 4 ways total.
We just enumerated all the possible cases, and all the possible cases that satisfy your probability, and we obtain 4/30. Note that I am summing 2+2 and then dividing by 30 in the same way that your method considers two equal probabilities. It is pure coincidence that they are equal.
A: Your answer is correct. It could also be done by realizing that when you choose 2 out of 6 marbles you must choose...


*

*0 out of 3 red marbles  

*1 out of 1 green marbles  

*1 out of 2 yellow
marbles


So $$ P(0,1,1)=  \frac{\binom 30 \binom 11 \binom 21 }{\binom 62} =\frac 2{16}$$
