Use the method of your choice to determine the following probability.
Selecting at random a green long-sleeved shirt from a closet in which five-sixths of the shirts are short-sleeved and one-sixth are long-sleeved and one-third of the long-sleeved shirts are green.
(Type an integer or a simplified fraction.)

Step 1 ：The problem is asking for the probability of selecting a green long-sleeved shirt. We know that one-sixth of the shirts are long-sleeved and one-third of the long-sleeved shirts are green.

Step 2 ：We can calculate this probability by multiplying these two probabilities together.

Step 3 ：Let's denote the probability of a shirt being long-sleeved as $P(L)=\frac{1}{6}$ and the probability of a long-sleeved shirt being green as $P(G|L)=\frac{1}{3}$.

Step 4 ：The probability of selecting a green long-sleeved shirt is then given by $P(G\cap L)=P(L)\times P(G|L)$.

Step 5 ：Substituting the given values, we get $P(G\cap L)=\frac{1}{6}\times \frac{1}{3}=\frac{1}{18}$.

Step 6 ：Final Answer: The probability of selecting a green long-sleeved shirt from the closet is $\boxed{{\displaystyle \frac{1}{18}}}$.