1. Key Concepts and Formulas

• Probability of Mutually Exclusive and Exhaustive Events: For a set of events that are mutually exclusive (cannot occur simultaneously) and exhaustive (cover all possible outcomes), the sum of their individual probabilities is 1. For a coin toss, $P(\text{Head}) + P(\text{Tail}) = 1$.
• Probability of Independent Events: If multiple events are independent (the outcome of one does not affect the others), the probability of all of them occurring in a specific sequence is the product of their individual probabilities. For events A and B, $P(A \text{ and } B) = P(A) \times P(B)$.
• Binomial Probability (Combinations and Arrangements): When finding the probability of a specific number of "successes" (e.g., tails) in a fixed number of independent trials, we must account for all possible orders in which these successes can occur. The formula for binomial probability is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success in a single trial.

2. Step-by-Step Solution

• Step 1: Determine the individual probabilities of getting a Head and a Tail in a single toss.

  • What we are doing: We need to find the numerical values for $P(H)$ and $P(T)$ for a single toss, considering the coin's bias.
  • Why we are doing this: These probabilities are fundamental for calculating the probabilities of sequences of multiple tosses.
  • Math: Let $P(T)$ be the probability of getting a tail in a single toss. The problem states that "a head is twice as likely to occur as a tail." Therefore, we can write: $P(H) = 2 \times P(T)$ Since a coin toss can only result in a head or a tail, these two events are mutually exclusive and exhaustive. Thus, their probabilities must sum to 1: $P(H) + P(T) = 1$ Now, substitute the expression for $P(H)$ into the sum equation: $(2 \times P(T)) + P(T) = 1$ $3 \times P(T) = 1$ $P(T) = \frac{1}{3}$ Now, we can find $P(H)$: $P(H) = 2 \times P(T) = 2 \times \frac{1}{3} = \frac{2}{3}$
  • Reasoning: By translating the problem's bias statement into an algebraic relationship and using the axiom of total probability, we uniquely determined the probabilities for each outcome of a single toss.

• Step 2: Identify all possible sequences for "two tails and one head" in three tosses.

  • What we are doing: We need to list all the distinct arrangements of two tails (T) and one head (H) when tossing a coin three times.
  • Why we are doing this: Each sequence represents a specific ordered outcome, and we need to calculate the probability for each of these before summing them up.
  • Math: Let H denote a Head and T denote a Tail. The possible sequences are:
    1. TTH (Tail, Tail, Head)
    2. THT (Tail, Head, Tail)
    3. HTT (Head, Tail, Tail) The number of such arrangements can also be found using combinations: $\binom{3}{2}$ ways to choose the positions for the two tails (or $\binom{3}{1}$ ways to choose the position for the one head), which equals $3$.
  • Reasoning: Listing these sequences ensures we consider all specific orders that fulfill the condition "two tails and one head."

• Step 3: Calculate the probability of each specific sequence.

  • What we are doing: For each sequence identified in Step 2, we will calculate its probability using the individual probabilities from Step 1.
  • Why we are doing this: Each coin toss is an independent event, so we can use the multiplication rule for independent probabilities.
  • Math: Using $P(T) = \frac{1}{3}$ and $P(H) = \frac{2}{3}$:
    1. For TTH: $P(\text{TTH}) = P(T) \times P(T) \times P(H) = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27}$
    2. For THT: $P(\text{THT}) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27}$
    3. For HTT: $P(\text{HTT}) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{2}{27}$
  • Reasoning: Due to the independence of each toss, the probability of a specific sequence is simply the product of the probabilities of the individual outcomes in that sequence. Note that all sequences with two tails and one head have the same probability.

• Step 4: Sum the probabilities of all favorable sequences.

  • What we are doing: We add the probabilities of the individual sequences calculated in Step 3.
  • Why we are doing this: Since the sequences (TTH, THT, HTT) are mutually exclusive (only one can occur in a specific set of three tosses), the total probability of getting two tails and one head is the sum of their individual probabilities.
  • Math: Total Probability $= P(\text{TTH}) + P(\text{THT}) + P(\text{HTT})$ Total Probability $= \frac{2}{27} + \frac{2}{27} + \frac{2}{27}$ Total Probability $= 3 \times \frac{2}{27} = \frac{6}{27}$ Simplify the fraction: Total Probability $= \frac{6 \div 3}{27 \div 3} = \frac{2}{9}$
  • Reasoning: The sum rule for mutually exclusive events allows us to combine the probabilities of all successful outcomes into a single overall probability.

3. Common Mistakes & Tips

• Common Mistake 1: Assuming a Fair Coin: Always pay close attention to whether a coin (or die, etc.) is biased. If it is, the first step must be to correctly determine the individual probabilities of each outcome ($P(H)$ and $P(T)$ here) before proceeding.
• Common Mistake 2: Forgetting Arrangements: A common error is to only calculate the probability of one specific sequence (e.g., $P(T) \times P(T) \times P(H) = 2/27$) and present that as the final answer. Remember that the order often doesn't matter unless specified, so you must account for all possible arrangements that satisfy the condition.
• Tip: Recognize Binomial Probability: This problem is a classic example of a binomial probability distribution. Once $P(T)$ and $P(H)$ are known, you can directly use the binomial probability formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Here, $n=3$ (tosses), let "success" be getting a tail, so $p=P(T)=1/3$. We want $k=2$ tails. $P(\text{2 Tails}) = \binom{3}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{3-2} = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9}$.

4. Summary

To determine the probability of getting two tails and one head in three tosses of a biased coin, we first established the probabilities of a single head ($2/3$) and a single tail ($1/3$) based on the given bias. Next, we identified all three possible sequences that result in two tails and one head (TTH, THT, HTT). We then calculated the probability of each sequence by multiplying the probabilities of its individual outcomes, leveraging the independence of each toss. Finally, by summing the probabilities of these mutually exclusive sequences, we arrived at the total probability of $2/9$.

5. Final Answer

The final answer is $\boxed{\frac{2}{9}}$ which corresponds to option (B).