A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Let the probability of $2$ appearing in a throw be given by $P\left(2\right)$ and the probability of $2$ not appearing be given by $P\left(2^{\text{'}}\right)$.

$\Rightarrow P\left(2\right)=\frac{1}{6}, P\left(2^{\text{'}}\right)=\frac{5}{6}$

Now, we want $2$ to appear in $2^{nd} \text{throw or} 4^{th} \text{throw or} 6^{th} \text{throw} ... \text{and so on}$

So, required probability is given by,

$P\left(E\right)=\frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^3\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^5\times \frac{1}{6}+...$

$\Rightarrow P\left(E\right)=\frac{1}{6}\times \frac{\frac{5}{6}}{1-\frac{25}{36}}$

$\Rightarrow P\left(E\right)=\frac{1}{6}\times \frac{\frac{5}{6}}{\frac{11}{36}}$

$\Rightarrow P\left(E\right)=\frac{5}{6}\times \frac{36}{11}\times \frac{1}{6}$

$\Rightarrow P\left(E\right)=\frac{5}{11}$