## Teaser 2889: Catching the bus

**From The Sunday Times, 4th February 2018** [link]

I can travel to town by either the number 12 bus or the number 15 bus. The number 12 bus leaves every 12 minutes and the number 15 bus leaves every 15 minutes. The first number 15 bus leaves soon after the first number 12 bus. I arrive at the bus stop at random and catch the first bus to arrive.

If the probability of me catching the number 15 bus is 2/5 how soon does the first number 15 bus leave after the first number 12 bus?

[teaser2889]

## Jim Randell 9:51 am

on12 December 2019 Permalink |(See also:

Teaser 2904).If we assume the #12 leaves at t=0, 12, 24, 36, 48, 60 minutes, and the first #15 bus leaves at

xminutes after the first #12 bus, then ifx < 3we have the following schedule:And the pattern repeats in subsequent hours.

If we arrive at a random time during the hour the probability of catching a #15 bus is:

Which means the first #15 bus leaves 1m30s after the first #12 bus.

However, if

3 < x < 6, then the #15 bus that leaves at(x + 45), departsafterthe #12 bus that departs at 45 minutes past the hour.We get:

Which means the first #15 bus leaves 4m30s after the first #12 bus.

And we also get workable values at 7m30s and 10m30s.

As found by this program which considers possible values (in seconds) for the times of the first #15 bus between the first and second #12 buses.

Run:[ @repl.it ]We are told that “the first #15 bus leaves soon after the first #12 bus”, so we can suppose we want the earliest possible value of 1m30s (although 4m30s would also seems to be “soon after” the first #12 bus).

Solution:The first number 15 bus leaves 1m30s after the first number 12 bus.If the first #15 bus does not have to leave before the second #12 bus then we can get solutions where the #15 buses can run on a timetable of integer minutes, and still satisfy the probability constraint given in the hour from the first #12 bus.

For instance if the first #15 bus leaves at 17m (so the timetable is: 17m, 32m, 47m, and then: 1h2m, 1h17m, 1h32m, 1h47m, …) then the probability of catching a #15 bus in the first hour is 0.4, similarly if the first #15 leaves at 29m.

Here is a graph of “the probability of catching a #15 bus in the first hour” (y-axis) against “time of departure of the first #15 bus” (x-axis).

We see that the probability is 0.4 for times of: 1m30s, 4m30s, 7m30s, 10m30s, 13m30s, 17m, 29m.

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