2018-II-43

Ans: C
The required probability

$\begin{array}{cl}
= & \dfrac{1}{6} + \dfrac{4}{6} \times \dfrac{4}{6} \times \dfrac{1}{6} + \dfrac{4}{6} \times \dfrac{4}{6} \times \dfrac{4}{6} \times \dfrac{4}{6} \times \dfrac{1}{6} + \cdots \\
= & \dfrac{\frac{1}{6}}{1 – \frac{4}{6} \times \frac{4}{6}} \\
= & \dfrac{3}{10}
\end{array}$