1. Find the probability of Head and Tail

Let the probability of tail be $x$.

Since a head is twice as likely as a tail, $P(H)=2x$

Also, $P(H)+P(T)=1$ So, $2x+x=1 \Rightarrow 3x=1 \Rightarrow x=\frac{1}{3}$

Hence, $P(T)=\frac{1}{3}, \qquad P(H)=\frac{2}{3}$

2. Find the probability of getting exactly two tails and one head in 3 tosses

We need outcomes with exactly 2 tails and 1 head.

The possible arrangements are:

• $TTH$
• $THT$
• $HTT$

So, number of favorable sequences is $\binom{3}{2}=3$

3. Probability of each such sequence

For example, probability of $TTH$ is $P(TTH)=P(T)\cdot P(T)\cdot P(H)=\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{27}$

Each of the 3 arrangements has the same probability $\frac{2}{27}$.

Therefore, total probability is $3\times \frac{2}{27}=\frac{6}{27}=\frac{2}{9}$

4. Match with the options

$\frac{2}{9}$ corresponds to Option B.

Therefore, the correct answer is B.