[ Discrete distribution ]
[Geometric progression ]

A and B are shooting at a range, but they only have a six-chambered 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.
A shot occurs on an odd attempt (when the revolver is at A), and no shot occurs on other attempts. Therefore, the desired probability is the sum $1 / 6+(5 / 6)^{2} \cdot 1 / 6+(5 / 6)^{4} \cdot 1 / 6+\ldots=6 / 11$.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_53798a1e86d520071242g-16.jpg?height=109&width=1172&top_left_y=183&top_left_x=0)
Difficulty: 3
Classes: $8,9,10,11$

100 bulbs were hung in a row on a New Year's tree. Then the bulbs started switching according to the following algorithm: all bulbs lit up, after a second every second bulb went out, after another second every third bulb switched: if it was lit, it went out and vice versa. After a second, every fourth bulb switched, after another second - every fifth, and so on. After 100 seconds, everything was over. Find the probability that a randomly chosen bulb is lit after this (the bulbs do not burn out or break).

## Solution

Obviously, a bulb with number $n$ will remain lit only if it has been switched an odd number of times, that is, if the number $n$ has an odd number of natural divisors. Clearly, this condition is satisfied only by squares: $n=1,4,9, \ldots, 100$ (see problem $\underline{30365}$ ). Thus, 10 bulbs out of 100 will remain lit. Therefore, the probability of randomly selecting a lit bulb is $0.1$.

## Answer

$0.1$.