91Ó°ÊÓ

Here, it is given that -

Trains headed for destination A arrive at the station at 15-minute intervals starting at $7$a.m.

Trains headed for destination B arrive at the station at 15-minute intervals starting at $7:05$ a.m.

If the passenger arrives at the station between $7$a.m. to $8$ a.m., then the probability that he or she will go to destination A is

role="math" localid="1646649389421" $$\begin{gathered} ={∫}_5^{15}\frac{1}{60}dx+{∫}_{20}^{30}\frac{1}{60}dx+{∫}_{35}^{45}\frac{1}{60}dx+{∫}_{50}^{60}\frac{1}{60}dx \\ =\left(\frac{15}{60}-\frac{5}{60}\right)+\left(\frac{30}{60}-\frac{20}{60}\right)+\left(\frac{45}{60}-\frac{35}{60}\right)+\left(\frac{60}{60}-\frac{50}{60}\right) \\ =\frac{10}{60}+\frac{10}{60}+\frac{10}{60}+\frac{10}{60} \\ =\frac{40}{60} \\ =\frac{2}{3} \end{gathered}$$

Therefore, the passenger takes the train to destination A two-thirds of the time.

If the passenger arrives at the station between $7:10$a.m. to $8:10$a.m., then the probability that he or she will go to destination A is

$$\begin{gathered} ={∫}_5^{15}\frac{1}{60}dx+{∫}_{20}^{30}\frac{1}{60}dx+{∫}_{35}^{45}\frac{1}{60}dx+{∫}_{50}^{60}\frac{1}{60}dx \\ =\left(\frac{15}{60}-\frac{5}{60}\right)+\left(\frac{30}{60}-\frac{20}{60}\right)+\left(\frac{45}{60}-\frac{35}{60}\right)+\left(\frac{60}{60}-\frac{50}{60}\right) \\ =\frac{10}{60}+\frac{10}{60}+\frac{10}{60}+\frac{10}{60} \\ =\frac{40}{60} \\ =\frac{2}{3} \end{gathered}$$

Therefore, the passenger takes the train to destination A two-thirds of the time.