A coin is biased such that the probability of getting heads is twice the probability of getting tails. If the coin is tossed twice, what is the probability of getting exactly one head?

Solution

1. Let $p$ be the probability of tails, then $2p$ is the probability of heads.
2. Since $p+2p=1$, we have $p=1/3$ and $P(H)=2/3$.
3. The outcomes for exactly one head are $HT$ and $TH$.
4. The probability is $$\begin{array}{rl}P(HT)+P(TH)& =(2/3\times 1/3)+(1/3\times 2/3)\\ & =2/9+2/9\\ & =4/9\end{array}$$.
5. Hence the answer is (C).