Let there be n throws in which a multiple of 3 occurs every time.
Probability of getting a multiple of 3 (i.e. 3 or 6 ) in one throw $=\frac{2}{6}=\frac{1}{3}$
$\therefore$ Probability of getting a multiple of 3 in $\mathrm{n}$ throws $=\left(\frac{1}{3}\right)^{\mathrm{n}}$
Probability of getting a 6 in one throw $=\frac{1}{6}$
$\therefore$ Probability of getting a 6 in $\mathrm{n}$ throws $=\left(\frac{1}{6}\right)^{\mathrm{n}}$
$\Rightarrow$ Probability of geetting at least a 3 in $\mathrm{n}$ throws $=\left(\frac{1}{3}\right)^{\mathrm{n}}-\left(\frac{1}{6}\right)^{\mathrm{n}}$
Let a multiple of 3 does not occur in $(\mathrm{n}+1)^{\text {th }}$ throw.
$\therefore$ Probability of getting $1,2,3,4,5$ (not a multiple of 3 ) in $(\mathrm{n}+1)^{\text {th }}$ throw $=\frac{4}{6}=\frac{2}{3}$
In the next throw a coin is tosses and tail occurs.
$\therefore$ Probabiltiy of getting a tail $=\frac{1}{2}$
$\Rightarrow$ Probability of getting at least a 3 and a tail in the end in $(x+2)^{\text {th }}$ throw.
$$
=\left[\left(\frac{1}{3}\right)^{\mathrm{n}}-\left(\frac{1}{6}\right)^{\mathrm{n}}\right] \frac{2}{3} \times \frac{1}{2}
$$
As $n \rightarrow \infty$; the probabiltiy of getting at least a 3 till tail is obtained.
$$
=\sum_{n=1}^{\infty}\left[\left(\frac{1}{3}\right)^n-\left(\frac{1}{6}\right)^n\right] \times \frac{2}{3} \times \frac{1}{2}
$$
$$
\begin{aligned}
&=\left[\frac{\frac{1}{3}}{1-\frac{1}{3}}-\frac{\frac{1}{6}}{1-\frac{1}{6}}\right] \times \frac{1}{3}=\left(\frac{1}{3} \times \frac{3}{2}-\frac{1}{6} \times \frac{6}{5}\right) \times \frac{1}{3} \\
&=\left(\frac{1}{2}-\frac{1}{5}\right) \times \frac{1}{3}=\frac{1}{3} \times \frac{3}{10}=\frac{1}{10}
\end{aligned}
$$