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Q. 5.10

Expert-verifiedFound in: Page 212

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 distributed

between $7:10$ 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.

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

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

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.

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