A fair die is thrown until 2 appears.

Question

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

Accepted Answer

Required probability $=$

$$\begin{aligned}
& \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \
& =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}
\end{aligned}$$

Solutions · 1