A coin is tossed three times. Let  $$X$$ denote the number of times a tail follows a head. If $$ \mu \text{ and } \sigma^{2} \text{denote the mean and variance of } X, \text{ then the value of }64(\mu+\sigma^{2}) \text{ is :} $$

To solve this problem, we need to find the mean $$\mu$$ and variance $$\sigma^2$$ of the random variable $$X$$, which denotes the number of times a tail follows a head in three tosses of a fair coin. Then, we compute $$64(\mu + \sigma^2)$$ and match it to the given options.

Since the coin is fair, each toss has two equally likely outcomes: head (H) or tail (T). With three tosses, the sample space has $$2^3 = 8$$ outcomes, each with probability $$\frac{1}{8}$$. We list all outcomes and determine $$X$$ for each by counting the number of times a tail immediately follows a head in consecutive tosses.

The outcomes and corresponding $$X$$ values are:

• HHH: Pairs (H,H) and (H,H) → No tail follows head → $$X = 0$$
• HHT: Pair (H,H) → No; Pair (H,T) → Yes → $$X = 1$$
• HTH: Pair (H,T) → Yes; Pair (T,H) → No → $$X = 1$$
• HTT: Pair (H,T) → Yes; Pair (T,T) → No → $$X = 1$$
• THH: Pair (T,H) → No; Pair (H,H) → No → $$X = 0$$
• THT: Pair (T,H) → No; Pair (H,T) → Yes → $$X = 1$$
• TTH: Pair (T,T) → No; Pair (T,H) → No → $$X = 0$$
• TTT: Pairs (T,T) and (T,T) → No → $$X = 0$$

Tabulating the values:

• $$X = 0$$ for HHH, THH, TTH, TTT → 4 outcomes
• $$X = 1$$ for HHT, HTH, HTT, THT → 4 outcomes

Note that $$X$$ can only be 0 or 1 because it is impossible to have two occurrences (as that would require overlapping pairs like (H,T) and (T,H), which conflict on the middle toss).

The probability mass function is:

• $$P(X=0) = \frac{4}{8} = \frac{1}{2}$$
• $$P(X=1) = \frac{4}{8} = \frac{1}{2}$$

The mean $$\mu$$ is the expected value of $$X$$. The formula for expected value is $$\mu = E(X) = \sum x_i \cdot P(X = x_i)$$.

So,

$$\mu = 0 \cdot P(X=0) + 1 \cdot P(X=1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

The variance $$\sigma^2$$ is given by $$\sigma^2 = E(X^2) - [E(X)]^2$$. First, compute $$E(X^2)$$:

$$E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

Then,

$$\sigma^2 = E(X^2) - [E(X)]^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Now, compute $$\mu + \sigma^2$$:

$$\mu + \sigma^2 = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$

Then, $$64(\mu + \sigma^2) = 64 \cdot \frac{3}{4} = 64 \cdot 0.75 = 48$$.

Thus, the value is 48, which corresponds to option D.