Problem (Combinatorics):
A bag contains some blue and some green hats. On each turn, Julia removes one hat without looking, with each hat in the bag being equally likely to be chosen. If it is green, she adds a blue hat into the bag from her supply of extra hats, and if it is blue, she adds a green hat to the bag. The bag initially contains 4 blue hats and 2 green hats. What is the probability that the bag again contains 4 blue hats and 2 green hats after two turns?

Solution:
Starting with 4 blue hats and 2 green hats, the probability that Julia removes a blue hat is $\frac{4}{6}=\frac{2}{3}$. The result would be 3 blue hats and 3 green hats, since a blue hat is replaced with a green hat.

In order to return to 4 blue hats and 2 green hats from 3 blue and 3 green, Julia would need remove a green hat (which would be replaced by a blue hat). The probability of her



removing a green hat from 3 blue and 3 green is $\frac{3}{6}=\frac{1}{2}$.

Summarizing, the probability of choosing a blue hat and then a green hat is $\frac{2}{3} \times \frac{1}{2}=\frac{1}{3}$.

Starting with 4 blue hats and 2 green hats, the probability that Julia removes a green hat is $\frac{2}{6}=\frac{1}{3}$. The result would be 5 blue hats and 1 green hat, since a green hat is replaced with a blue hat.

In order to return to 4 blue hats and 2 green hats from 5 blue and 1 green, Julia would need remove a blue hat (which would be replaced by a green hat). The probability of her removing a green hat from 5 blue and 1 green is $\frac{5}{6}$.

Summarizing, the probability of choosing a green hat and then a blue hat is $\frac{1}{3} \times \frac{5}{6}=\frac{5}{18}$.

These are the only two ways to return to 4 blue hats and 2 green hats after two turns removing a blue hat then a green, or removing a green then a blue.

Therefore, the total probability of returning to 4 blue hats and 2 green hats after two turns is $\frac{1}{3}+\frac{5}{18}=\frac{11}{18}$.

Final answer: $\frac{11}{18}$