Example 6.26. A master, who performs repairs at home, can appear at any time from 10 to 18 hours. The client, having waited until 14 o'clock, left for one hour. What is the probability that the master (his arrival is mandatory) will not find him at home?
Solution. From the analysis of the problem statement, we accept the uniform distribution law for the random arrival time of the master. To determine the boundaries of the time interval, we note that since the master did not arrive by 14:00, he will definitely arrive by 18:00. Therefore, \(a=14\) and \(b=18\), and the cumulative distribution function is:

$$
F(x)=\frac{x-a}{b-a}=\frac{x-14}{18-14}
$$

Then the probability of his appearance during the client's absence (from 14:00 to 15:00) is:

$$
P(14<X<15)=F(15)-F(14)=\frac{15-14}{18-14}-\frac{14-14}{18-14}=\frac{1}{4} .
$$