Explanation

(b)
Let $E$ be the event of getting 1 on a die
$\Rightarrow P(E) = \frac{1}{6}\ and\ P\left( \overline{E} \right) = \frac{5}{6}$
$\therefore P$(first time 1 occurs at the even throw)
$= t_{2}\ or\ t_{4}\ or{\ t}_{6}\ or\ t_{8}\ldots$ and so on.
$= \left\{ P\left( {\overline{E}}_{1} \right).P\left( E_{2} \right) \right\} + \left\{ P\left( {\overline{E}}_{1} \right)P\left( {\overline{E}}_{2} \right)P\left( \overline{E_{3}} \right)P\left( E_{4} \right) \right\} + ...\infty$
$= \left( \frac{5}{6}.\frac{1}{6} \right) + \left( \frac{5}{6} \right)^{3}\left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^{3}\left( \frac{1}{6} \right) + ...\infty$
$= \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{5}{11}$