In each bag there are red, green, and blue tiles, meaning that no matter which color is pulled out first there is always some probability that the second tile will be the same color. So, we can set up three possible outcomes:

Red: The player pulls out a red tile first. This has a $\frac{1}{9}$ probability of happening. Then in order to succeed for the problem, the next tile also needs to be red which has a $\frac{6}{12}$ probability attached to it. $\frac{1}{9}$ × $\frac{6}{12}$ = $\frac{1}{18}$ probability of happening.

Green: There is a $\frac{5}{9}$ probability of the player pulling out a green tile first. In this case we want to calculate the probability of the second tile being green, which would be $\frac{4}{12}$ . $\frac{5}{9}$ × $\frac{4}{12}$ = $\frac{5}{27}$ .

Blue: There is a $\frac{3}{9}$ probability of the first tile being blue in which case we are hoping for the second tile to be blue as well. The probability of the second tile being blue is $\frac{2}{12}$ on its own, and them both being blue is $\frac{3}{9}$ × $\frac{2}{12}$ = $\frac{1}{18}$

Adding $\frac{1}{18}$ + $\frac{1}{18}$ + $\frac{5}{27}$ we get the answer $\frac{8}{27}$ .

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3 years ago