Péter's favorite song is the eighth track on his CD, which contains twelve tracks in total. When he puts the disc into the player, he starts with the first track by pressing a button once, and then it takes him seven more button presses to get to his favorite song. If the device is in "random" mode, it will play the 11 tracks in a randomly shuffled order. What is the probability that he will reach his favorite song with fewer button presses in this mode?
I. solution. In normal operation mode, Peter reaches his favorite number with a total of eight button presses. Let $x$ denote the number of button presses with which Peter reaches his favorite (eighth) number in "random" mode ( $x \in \mathbb{N}$ ).

The probability that $x<8$ :

$$
\begin{aligned}
p(x<8)= & p(x=1)+p(x=2)+p(x=3)+\ldots+p(x=7)= \\
= & \frac{1}{11}+\frac{10}{11} \cdot \frac{1}{10}+\frac{10}{11} \cdot \frac{9}{10} \cdot \frac{1}{9}+\ldots+\frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{1}{6}+ \\
& +\frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{1}{5}=7 \cdot \frac{1}{11}=\frac{7}{11}
\end{aligned}
$$

Thus, the probability that Peter reaches his favorite number in fewer than eight button presses in "random" mode is: $\frac{7}{11}$.

Note. A program I created reached the eighth number in fewer than eight "button presses" in $63,63814 \%$ of 10 million runs. Indeed: $\frac{7}{11}=0, \dot{6} \dot{3} \approx 63.64 \%$.

II. solution. There are a total of 11! possible arrangements. After the shuffle, the favorite song can be in any of the first 7 positions. In each of the 7 cases, the possible arrangements of the other numbers are 10!. Thus, the probability is:

$$
\frac{7 \cdot 10!}{11!}=\frac{7}{11}
$$