16. Technical Break (recommended for 9th grade, 3 points). A woman working as a cashier at a railway ticket office saw that there were no customers in front of the window, sighed, hung a sign saying "Technical Break 15 minutes," and left for exactly 15 minutes. When the Absent-Minded Scientist approached the ticket window, he saw that there were already five people in line, waiting patiently for the window to open. What is the probability that the ticket office will open no later than 3 minutes after the Scientist's arrival?
Solution. We will measure time in minutes from the moment the cashier leaves. Let $x_{1}, x_{2}, \ldots, x_{5}, x_{6}$ be the moments when five people and the Scholar approached the window. All these numbers are greater than zero but less than 15, and it is known that the sixth number is greater than the others. We will briefly write this as $x_{1}, x_{2}, \ldots, x_{5}<x_{6}$ and consider it as event $B$.

Given this condition, we need to find the probability of event $A=\left(x_{6} \geq 12\right)$. We will find the probability of the opposite event

$$
\mathrm{P}(\bar{A} \mid B)=\mathrm{P}(\bar{A} \cap B \mid B)=\mathrm{P}\left(x_{1}, \ldots, x_{5}, x_{6}<12 \mid B\right)
$$

Events $x_{1}, \ldots, x_{5}, x_{6}<12$ "all six numbers are less than 12" and $B$ "the sixth number is greater than the others" are independent, since under the condition "all numbers are less than 12," any of them can be the largest with equal probability. Therefore,

$$
\mathrm{P}(\bar{A} \mid B)=\mathrm{P}\left(x_{1}, \ldots, x_{5}, x_{6}<12\right)=\left(\frac{12}{15}\right)^{6}=0.8^{6}
$$

The desired probability is $\mathrm{P}(A \mid B)=1-0.8^{6} \approx 0.738$.

Answer: 0.738.