Q. 5.10 - A First Course in Probability

Question

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

(a) If a certain passenger arrives at the station at a time uniformly distributed between $7$ and $8$ a.m. and then gets on the first train that arrives, what proportion of time does he or she go to destination A?

(b)What if the passenger arrives at a time uniformly distributed

between $7:10$ and $8:10$ a.m.?

Step-by-Step Solution

Answer

(a) If a passenger arrives at the station at a time uniformly distributed between $7$ and $8$ a.m. and then gets on the first train that arrives, then for $\frac{2}{3}$times he or she goes to destination A

(b) If the passenger arrives at a time uniformly distributed between $7:10$ and $8:10$ a.m., then for $\frac{2}{3}$times he or she goes to destination A.

Part (a) Step 1. Given information.

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. 

Part (a) Step 2. Calculate the proportion of time the passenger goes to destination A, if he or she arrives the station between 7 am to 8 am.

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

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

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

Part (b) Step 1. Calculate the proportion of time the passenger goes to destination A, if he or she arrives the station between 7 : 10 a.m. to 8 : 10 a.m.

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

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

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