To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).

Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $$ P(T) = p $$ and the probability of getting a head as $$ P(H) = 2p $$.

These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :

$$ P(H) + P(T) = 1 $$

$$ 2p + p = 1 $$

$$ 3p = 1 $$

$$ p = \frac{1}{3} $$

Therefore, the probability of getting a tail (T) is $$ P(T) = \frac{1}{3} $$ and the probability of getting a head (H) is $$ P(H) = 2 \times \frac{1}{3} = \frac{2}{3} $$.

Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur. There are three unique sequences: TTH, THT, and HTT.

The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:

$$ P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27} $$

$$ P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} $$

$$ P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27} $$

The overall probability of getting two tails and one head in any order is the sum of these individual probabilities :

$$ P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27} $$

Simplifying this expression gives us: $$ P(2T1H) = \frac{6}{27} = \frac{2}{9} $$

Therefore, the correct answer is :

Option B : $$\frac{2}{9}$$