Here, two events [throwing the die denoted as $\mathrm{D}$ first and then tossing the coin denoted as $\mathrm{C}$] should be occurred together. Because any odd number of events like $\text{D}$, $\text{DCD}$ or $\text{DCDCD}$ will never produce the desirable outcome. So the number of events will always be even. So the even-sequence will be $$\text{DC}, \text{DCDC}, \text{DCDCDC},\cdots \tag{i}$$
The last event $\mathrm{C}$ will always be $\mathrm{T}$ (Tail). Any $\mathrm{D}$ can be the number $5$ with a condition that the following $\mathrm{C}$ will NOT be $\mathrm{T}$ (Tail) until the last $\mathrm{C}$. Besides, if any middle $\mathrm{C}$ is $\mathrm{T}$, the previous $\mathrm{D}$ can be $\{1,2,3,4,6\}$ (i.e. $1$ to $6$ except $5$).
So for $\text{DCDC}$, we can have cases in the form $\mathrm{5H\_T}$ where $\_$ can be replaced by any from $\{1,2,3,4,5,6\}$ [i.e. $6$ cases] and $\mathrm{{*}T5T}$ where $*$ can be replaced by any from $\{1,2,3,4,6\}$ [i.e. $5$ cases].
Now
| Events |
Form of Cases |
Probability |
| $\text{DC}$ |
$\mathrm{5T}$ |
$\frac{1}{6\times2}=\frac{1}{12}$ |
| $\text{DCDC}$ |
$\mathrm{5H\_T}$, $\mathrm{{*}T5T}$ |
$\frac{6+5}{6\times2\times6\times2}=\frac{6+5}{12^2}$ |
| $\text{DCDCDC}$ |
$\mathrm{5H\_H\_T}$, $\mathrm{{*}T5H\_T}$, $\mathrm{{*}T{*}T5T}$
|
$\frac{6^2+5\times6+5^2}{6\times2\times6\times2\times6\times2}=\frac{5^06^2+5^16^1+5^26^0}{12^3}$ |
| $\text{DCDCDCDC}$ |
$\mathrm{5H\_H\_H\_T}$, $\mathrm{{*}T5H\_H\_T}$, $\mathrm{{*}T{*}T5H\_T}$, $\mathrm{{*}T{*}T{*}T5T}$ |
$\frac{6^3+5\times6^2+5^2\times6+5^3}{6\times2\times6\times2\times6\times2\times6\times2}=\frac{5^06^3+5^16^2+5^26^1+5^36^0}{12^4}$ |
| $\cdots$ |
$\cdots$ |
$\cdots$ |
$\therefore$ The probability of the $n^\mathrm{th}$ term of the sequence no$\mathrm{(i)}$ is
$\begin{align} P_n&=\frac{5^06^{n-1}+5^16^{n-2}+5^26^{n-3}+\cdots+5^{n-1}6^0}{12^n}\\&=\frac{6^{n-1}}{12^n} \left\{ 1+\left( \frac{5}{6} \right)+\left( \frac{5}{6} \right)^2+\left( \frac{5}{6} \right)^3+\cdots+\left( \frac{5}{6} \right)^{n-1} \right\}\\&=\frac{1}{6}\left( \frac{1}{2}\right)^n \left\{ \frac{1-\left( \frac{5}{6} \right)^n}{1-\left( \frac{5}{6} \right)} \right\}\\&=\left( \frac{1}{2}\right)^n \left\{1- \left(\frac{5}{6}\right)^n \right\}\\&=\left( \frac{1}{2}\right)^n-\left( \frac{5}{12}\right)^n\end{align}$
$\therefore$ The required probability
$\begin{align}&=P_1+P_2+P_3+\cdots\infty\\&= \left\{\left( \frac{1}{2}\right)^1-\left( \frac{5}{12}\right)^1\right\}+\left\{\left( \frac{1}{2}\right)^2-\left( \frac{5}{12}\right)^2\right\}+\left\{\left( \frac{1}{2}\right)^3-\left( \frac{5}{12}\right)^3\right\}+\cdots\infty\\&=\left\{\left( \frac{1}{2}\right)^1+\left( \frac{1}{2}\right)^2+\left( \frac{1}{2}\right)^3+\cdots\infty \right\}-\left\{\left( \frac{5}{12}\right)^1+\left( \frac{5}{12}\right)^2+\left( \frac{5}{12}\right)^3+\cdots\infty \right\}\\&=\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{5}{12}}{1-\frac{5}{12}}\\&=1- \frac{5}{7}\\&=\frac{2}{7}\end{align}$
So the correct answer is D.
