 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q. 5.10

Expert-verified Found in: Page 212 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # 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 distributedbetween $7:10$ and $8:10$ a.m.?

(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.

See the step by step solution

## 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 7am 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

role="math" localid="1646649389421" $={\int }_{5}^{15}\frac{1}{60}dx+{\int }_{20}^{30}\frac{1}{60}dx+{\int }_{35}^{45}\frac{1}{60}dx+{\int }_{50}^{60}\frac{1}{60}dx\phantom{\rule{0ex}{0ex}}=\left(\frac{15}{60}-\frac{5}{60}\right)+\left(\frac{30}{60}-\frac{20}{60}\right)+\left(\frac{45}{60}-\frac{35}{60}\right)+\left(\frac{60}{60}-\frac{50}{60}\right)\phantom{\rule{0ex}{0ex}}=\frac{10}{60}+\frac{10}{60}+\frac{10}{60}+\frac{10}{60}\phantom{\rule{0ex}{0ex}}=\frac{40}{60}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$

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 8:10a.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

$={\int }_{5}^{15}\frac{1}{60}dx+{\int }_{20}^{30}\frac{1}{60}dx+{\int }_{35}^{45}\frac{1}{60}dx+{\int }_{50}^{60}\frac{1}{60}dx\phantom{\rule{0ex}{0ex}}=\left(\frac{15}{60}-\frac{5}{60}\right)+\left(\frac{30}{60}-\frac{20}{60}\right)+\left(\frac{45}{60}-\frac{35}{60}\right)+\left(\frac{60}{60}-\frac{50}{60}\right)\phantom{\rule{0ex}{0ex}}=\frac{10}{60}+\frac{10}{60}+\frac{10}{60}+\frac{10}{60}\phantom{\rule{0ex}{0ex}}=\frac{40}{60}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$

Therefore, the passenger takes the train to destination A two-thirds of the time. ### Want to see more solutions like these? 