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

<table border="1"> <tr> <th>Outcome</th> <th>$x_i$</th> <th>$p_i$</th> </tr> <tr> <td>HHH</td> <td>0</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>TTT</td> <td>0</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>HHT</td> <td>1</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>HTH</td> <td>1</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>THH</td> <td>0</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>TTH</td> <td>0</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>THT</td> <td>1</td> <td>$\frac{1}{8}$</td> </tr> <tr> <td>HTT</td> <td>1</td> <td>$\frac{1}{8}$</td> </tr> </table> $\begin{aligned} & \mu=\sum x_i P_i=\frac{1}{2} \\ & \sigma^2=\sum x_i^2 P_i-\mu^2 \\ & =\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\ & 64\left(\mu+\sigma^2\right)=64\left[\frac{1}{2}+\frac{1}{4}\right] \\ & =64 \times \frac{3}{4}=48\end{aligned}$