Probability Part 2

Let $p = \frac{1}{6}$ be the probability of getting a 2 on a single throw, and $q = 1 - p = \frac{5}{6}$ be the probability of not getting a 2. Since the die is thrown until 2 appears, 2 appears for the first time on the $n$-th throw with probability $q^{\,n-1}p$. We need 2 to appear in an even number of throws, so $n = 2, 4, 6, \dots$ Hence the required probability is $$P = qp + q^3p + q^5p + \cdots = qp\left(1 + q^2 + q^4 + \cdots\right) = \frac{qp}{1 - q^2}$$ Substituting $p = \frac{1}{6}$ and $q = \frac{5}{6}$, $$P = \frac{\frac{5}{6} \cdot \frac{1}{6}}{1 - \left(\frac{5}{6}\right)^2} = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}$$ Therefore the probability that 2 appears in an even number of throws is $\frac{5}{11}$, so option C is correct. (The solution shown earlier belongs to a different complex-number question and is not relevant here.)