Question

Trains headed for destination A arrive at the train station at 15-minute intervals starting at 7:00 am, whereas trains headed for destination B arrive at 12-minute intervals starting at 7:07 am. If a passenger arrives at the station at a time uniformly distributed between 7:00 am and 8:00 am and then gets on the first train that arrives, what proportion of time does the passenger go to destination A?
a. 0.43
b. none of the answers provided here
c. 0.53
d. 0.67
e. 0.37

Answer 1

Answer:

The answer to the question is 0.43. Let's explain why.

Step-by-step explanation:

A train arrives at the station at 15-minute intervals starting from 7:00 am which makes the train arrives at;

7:00

7:15

7:30

7:45

8:00

B train arrive at 12-minute intervals starting at 7:07 means B train arrives at;

7:07

7:19

7:31

7:43

7:55

Now, let's review the chances of a passenger arrives at the station any time between 7:00 am and 8:00 am

• If the passenger arrives between 7:00 - 7:07 the passenger will get into B train which gives us the $\frac{7}{60}$ chance since an hour is 60 minutes and the passenger waited for 7 minutes.
• If the passanger arrives between 7:07 - 7:15 passenger will get into A train which gives $\frac{8}{60}$ chance. If we apply this to the time limits it goes by;
• Between 7:15 - 7:19 passenger will go to B destination which gives $\frac{4}{60}$ chance
• Between 7:19 - 7:30 passenger will go to A destination means $\frac{11}{60}$ chance
• Between 7:30 - 7:43 passenger will go to B destination means $\frac{13}{60}$ chance
• Between 7:43 - 7:45 passenger will go to A destination which gives $\frac{2}{60}$ chance
• Between 7:45 - 7:55 passenger will go to B destination means $\frac{10}{60}$ chance
• An lastly between 7:55 - 8:00 passenger will go to A destination which gives us $\frac{5}{60}$ chance

Now for passenger to go to A destination the total chances are:

$\frac{8}{60} + \frac{11}{60} + \frac{2}{60} + \frac{5}{60} = \frac{26}{60}$ = 0.43