A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is

<p>To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).</p> <p>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 $P(H) = 2p$.</p> <p>These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss : <br/><br/>$P(H) + P(T) = 1$ <br/><br/>$2p + p = 1$ <br/><br/>$3p = 1$ <br/><br/>$p = \frac{1}{3}$</p> <p>Therefore, the probability of getting a tail (T) is $P(T) = \frac{1}{3}$ and the probability of getting a head (H) is $P(H) = 2 \times \frac{1}{3} = \frac{2}{3}$.</p> <p>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.</p> <p>The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent: <br/><br/>$$ P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27} $$ <br/><br/>$$ P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} $$ <br/><br/>$$ P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27} $$</p> <p>The overall probability of getting two tails and one head in any order is the sum of these individual probabilities : <br/><br/>$$ P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27} $$</p> <p>Simplifying this expression gives us: $P(2T1H) = \frac{6}{27} = \frac{2}{9}$</p> <p>Therefore, the correct answer is : <br/><br/>Option B : $\frac{2}{9}$</p>