1. $(1 \mathbf{1}, 7-9)$ A and B are shooting at a shooting range, but they only have one six-chamber revolver with one bullet. Therefore, they agreed to take turns randomly spinning the cylinder and shooting. A starts. Find the probability that the shot will occur when the revolver is with A.
# Solution.

The shot occurs on an odd attempt (when the revolver is with A), and no shot occurs on the other attempts. Therefore, the desired probability is the sum

$$
\frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\ldots=\frac{1}{6}+\frac{1}{6} \cdot \frac{25}{36}+\frac{1}{6} \cdot\left(\frac{25}{36}\right)^{2}+\ldots=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}
$$

Answer: $\frac{6}{11}$.