1. Key Concepts and Formulas

   • Probability of Events: For a biased coin, if Head (H) is $k$ times as likely as Tail (T), then $P(H) = k \cdot P(T)$. Since H and T are the only outcomes, $P(H) + P(T) = 1$.
   • Expected Value (Mean) of a Discrete Random Variable: For a discrete random variable $X$ that can take values $x_1, x_2, \ldots, x_n$ with probabilities $P(X=x_1), P(X=x_2), \ldots, P(X=x_n)$ respectively, its expected value $E(X)$ is given by: $E(X) = \sum_{i=1}^{n} x_i P(X=x_i)$
   • Independent Events: Each coin toss is an independent event. The probability of a sequence of outcomes (e.g., TH) is the product of the probabilities of the individual outcomes in that sequence (e.g., $P(T) \cdot P(H)$).

2. Step-by-Step Solution

   Step 1: Determine the Probabilities of Head ($P(H)$) and Tail ($P(T)$) for a Single Toss

   The problem states that a head is 3 times as likely to occur as a tail. Let $P(H)$ be the probability of getting a Head and $P(T)$ be the probability of getting a Tail. From the problem statement: $P(H) = 3 \cdot P(T)$ We also know that the sum of probabilities of all possible outcomes must be 1: $P(H) + P(T) = 1$ Now, substitute the first equation into the second: $(3 \cdot P(T)) + P(T) = 1$ $4 \cdot P(T) = 1$ $P(T) = \frac{1}{4}$ Substitute $P(T)$ back into the equation for $P(H)$: $P(H) = 3 \cdot \left(\frac{1}{4}\right) = \frac{3}{4}$ So, the probabilities for a single toss are $P(H) = \frac{3}{4}$ and $P(T) = \frac{1}{4}$.

   Step 2: Identify All Possible Values of the Random Variable $X$ and Their Corresponding Sequences

   The coin is tossed until one of two stopping conditions is met:

   1. A Head (H) occurs.
   2. Three Tails (TTT) occur.

   We list the possible sequences of tosses and the number of tosses ($X$) until a stopping condition is met:

   • If $X=1$: The first toss is a Head. Sequence: H (Condition 1 met)

   • If $X=2$: The first toss must not be a Head (so it's a Tail), and the second toss must be a Head. Sequence: TH (Condition 1 met)

   • If $X=3$: The first two tosses must not have resulted in a Head (so they are TT). On the third toss, either a Head occurs or a Tail occurs. Sequence: TTH (Condition 1 met) Sequence: TTT (Condition 2 met)

   No other values for $X$ are possible. If the process hasn't stopped by $X=2$, the sequence must be TT. On the third toss, either TTH or TTT occurs, both of which are stopping conditions. Thus, $X$ can only take values $1, 2,$ or $3$.

   Step 3: Calculate the Probability for Each Possible Value of $X$

   Using $P(H) = \frac{3}{4}$ and $P(T) = \frac{1}{4}$, and the independence of tosses:

   • For $X=1$ (Sequence H): $P(X=1) = P(H) = \frac{3}{4}$

   • For $X=2$ (Sequence TH): $P(X=2) = P(T) \cdot P(H) = \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16}$

   • For $X=3$ (Sequences TTH or TTT): We sum the probabilities of these two mutually exclusive sequences: $P(\text{TTH}) = P(T) \cdot P(T) \cdot P(H) = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{64}$ $P(\text{TTT}) = P(T) \cdot P(T) \cdot P(T) = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{64}$ $P(X=3) = P(\text{TTH}) + P(\text{TTT}) = \frac{3}{64} + \frac{1}{64} = \frac{4}{64} = \frac{1}{16}$

   Let's verify that the sum of all probabilities is 1: $P(X=1) + P(X=2) + P(X=3) = \frac{3}{4} + \frac{3}{16} + \frac{1}{16} = \frac{12}{16} + \frac{3}{16} + \frac{1}{16} = \frac{12+3+1}{16} = \frac{16}{16} = 1$ The probabilities are correct.

   Step 4: Calculate the Expected Value (Mean) of $X$

   Using the formula $E(X) = \sum x_i P(X=x_i)$: $E(X) = (1 \cdot P(X=1)) + (2 \cdot P(X=2)) + (3 \cdot P(X=3))$ Substitute the values from Step 3: $E(X) = \left(1 \cdot \frac{3}{4}\right) + \left(2 \cdot \frac{3}{16}\right) + \left(3 \cdot \frac{1}{16}\right)$ $E(X) = \frac{3}{4} + \frac{6}{16} + \frac{3}{16}$ To sum these fractions, find a common denominator, which is 16: $E(X) = \frac{3 \cdot 4}{4 \cdot 4} + \frac{6}{16} + \frac{3}{16}$ $E(X) = \frac{12}{16} + \frac{6}{16} + \frac{3}{16}$ $E(X) = \frac{12 + 6 + 3}{16}$ $E(X) = \frac{21}{16}$

3. Common Mistakes & Tips

   • Misinterpreting Stopping Conditions: Ensure you correctly identify when the process stops. "Until a head OR three tails occur" means the first occurrence of either condition stops the experiment.
   • Missing Outcomes: Systematically list all possible sequences for each value of $X$ to avoid missing any valid paths. A good check is to ensure the sum of all $P(X=x_i)$ equals 1.
   • Incorrect Probability Calculation: Remember that for independent events, the probability of a sequence is the product of individual event probabilities. If multiple sequences lead to the same $X$ value, sum their probabilities.

4. Summary

   To find the mean (expected value) of $X$, the number of tosses, we first determined the individual probabilities of getting a head ($P(H) = 3/4$) and a tail ($P(T) = 1/4$) from the given bias. Next, we identified all possible values $X$ could take based on the stopping conditions (H or TTT), which were $X=1, 2, 3$. We then calculated the probability for each of these $X$ values: $P(X=1) = 3/4$, $P(X=2) = 3/16$, and $P(X=3) = 1/16$. Finally, we applied the expected value formula $E(X) = \sum x_i P(X=x_i)$ to sum the products of each $X$ value and its probability, arriving at the mean of $X$ as $\frac{21}{16}$.

The final answer is $\boxed{\frac{21}{16}}$, which corresponds to option (C).