A coin is biased so that a head is twice as likely to occur as a...

Given: $P\left(H\right) = 2P\left(T\right)$

We know that, $P\left(H\right) + P\left(T\right) = 1$

$⇒ 2P\left(T\right) + P\left(T\right) = 1$

$⇒ P\left(T\right) = \dfrac{1}{3}\text{and}P\left(H\right) = \dfrac{2}{3}$

$⇒ P\left(2\text{tails and}1\text{head}\right) = \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{3 !}{2 !}\left(\text{for arrangements}\right)$

$⇒ P\left(2\text{tails and}1\text{head}\right) = \dfrac{2}{9}$