Solution. The probability that no numbers from the first draw will be repeated in the second draw is $a=\frac{39 \cdot 38 \cdot 37 \cdot \ldots \cdot 34}{45 \cdot 44 \cdot 43 \cdot \ldots 40}=0.40056 \ldots$, which is slightly more than 0.4. The probability that there will be no repetitions from the previous draw in both the second and third draws is the square of this number, that is, $a^{2}$. The probability of no repetitions in the draws from the second to the fourth is $a^{3}$, and so on. The fifth power of the number $a$ is greater than 0.01: $a^{5}>0.4^{5}=10^{-5} \cdot 2^{10}=0.01024$, while the sixth is already less: $a^{6}=0.0041 \ldots$. Therefore, Pinocchio can accuse Karabas-Barabas after this phenomenon repeats 6 times, that is, after the seventh draw.

## Answer: after the seventh.
9. Buratino the Statistician (from 7th grade, 2 points). Every month, Buratino plays in the "6 out of 45" lottery organized by Karabas-Barabas. In the lottery, there are 45 numbered balls, and in each draw, 6 random winning balls are drawn.

Buratino noticed that in each subsequent draw, there are no balls that appeared in the previous draw: in the second draw, there were no balls from the first, in the third draw, there were no balls from the second, and so on.

Buratino does not believe in events with a probability of less than 0.01. If such an event occurs, Buratino begins to suspect something is amiss. After which draw will Buratino start to suspect that Karabas-Barabas is cheating?