Once in the autumn, the Absent-Minded Scholar glanced at his old wall clock and saw that three flies had fallen asleep on the dial. The first was sleeping precisely at the 12 o'clock mark, while the other two were neatly positioned at the 2 o'clock and 5 o'clock marks. The Scholar made some measurements and determined that the hour hand posed no threat to the flies, but the minute hand would sweep them all away one by one. Find the probability that exactly 40 minutes after the Scholar noticed the flies, exactly two out of the three flies were swept away by the minute hand.
We will call the flies - fly 2, fly 5, and fly 12 based on the position where they fell asleep.

The hand could have swept away fly 12 and fly 2 but not touch fly 5 only if at the moment of the first observation it was located between 6 and 9 o'clock, which gives 1/4 of the circle.

The hand could have swept away only fly 2 and fly 5 if it was located between 12 and 2 o'clock at the moment of the first observation. This gives 1/6 of the circle.

Finally, the hand could have swept away only flies 5 and 12 if initially it was located between 4 and 5 o'clock, which is a segment occupying $1 / 12$ of the circle. Thus, considering all initial positions of the hand equally possible and taking into account the uniformity of its movement, we get that the desired probability is $1 / 4 + 1 / 6 + 1 / 12 = 1 / 2$.

## Answer

A ticket for the train costs 50 rubles, and the fine for traveling without a ticket is 450 rubles. If a fare dodger (rabbit) is caught by a conductor, they have to pay both the fine and the ticket cost. It is known that a conductor is encountered on average once every 10 trips. The rabbit has familiarized himself with the basics of probability theory and decided to follow a strategy that minimizes the expected expenses. What should he do: buy a ticket every time, never buy one, or flip a coin to decide whether to buy a ticket or not?

## Solution

Let the probability of buying a ticket be $p$, and the probability of the conductor appearing be $q$. The random variable $X$ "Expenses for the trip" has the distribution

$$
\left(\begin{array}{ccc}
0 & 50 & 500 \\
(1-p)(1-q) & p & (1-p) q
\end{array}\right)
$$

The expected value of expenses $\mathrm{E} X=50 p+500(1-p) q=500 q-50 p(10 q-1)$.

According to the condition, $q=0.1$. Therefore, $\mathrm{E} X=50$, which means it does not depend on $p$. An "equilibrium" arises - the average expense per trip does not depend on the frequency or probability of buying a ticket.

## Answer

It doesn't matter.

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