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

Let $$p$$ be the probability of getting a $$2$$ in a single throw: $$p = \frac{1}{6}$$.

 Let $$q$$ be the probability of not getting a $$2$$: $$q = 1 - p = \frac{5}{6}$$.

The event "2 appears in an even number of throws" happens if it occurs on the 2nd, 4th, 6th... throw.

The total probability $$P$$ is an infinite geometric series:

$$P = (q \cdot p) + (q^3 \cdot p) + (q^5 \cdot p) + \dots$$

$$P = qp(1 + q^2 + q^4 + \dots)$$

Using the sum of an infinite GP $$S = \frac{a}{1-r}$$:

$$P = \frac{qp}{1 - q^2} = \frac{\frac{5}{6} \cdot \frac{1}{6}}{1 - (\frac{5}{6})^2} = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}$$