Standing at a bus stop where two buses stop. One once per five minutes, the other once per eight minutes. Find the probability that the bus that comes once per five minutes is the next bus to come.

Follow-up; as above, but three are three buses; one goes every three minutes, one every five minutes, and the third every eight minutes. Again, find the probability that the one to come next is the bus that goes once per five minutes.

The schedules of the different bus lines are independent of one another.

Originally posted by talzamir Standing at a bus stop where two buses stop. One once per five minutes, the other once per eight minutes. Find the probability that the bus that comes once per five minutes is the next bus to come.

Follow-up; as above, but three are three buses; one goes every three minutes, one every five minutes, and the third every eight minutes. Again, find the proba ...[text shortened]... e per five minutes.

The schedules of the different bus lines are independent of one another.

From any given time, either bus has an equally distributed probability curve that the bus will come in the next n minutes, where n is less than that busses cycle time (c)

P(arrival < n) = n / c

Therefore, the probability that the 8min bus will come in the next 5 minutes is 5/8

If the 8min bus will come in the next 5 min, I'm pretty sure that there is an equal likelihood of either bus arriving first. That leaves the probability that the 8min bus will arrive first at Reveal Hidden Content

(1/2)*(5/8) = 5/16

Therefore, the probability that the 5min bus will arrive first is Reveal Hidden Content

Originally posted by forkedknight From any given time, either bus has an equally distributed probability curve that the bus will come in the next n minutes, where n is less than that busses cycle time (c)

P(arrival < n) = n / c

Therefore, the probability that the 8min bus will come in the next 5 minutes is 5/8

If the 8min bus will come in the next 5 min, I'm pretty sure that th ...[text shortened]... 8.75%[/hidden]

For part 2, I get [hidden]1/3 * 3/5 * 3/8 + 1/2 * 3/5 * 5/8 = 26.25%[/hidden]

Ah I was missing part. I'm not well aquainted with probability.

For continuous distribution I sometimes like to think things in terms of geometry. For the first problem it works nicely - drawing a rectangle, five minutes by eight minutes, where every point refers to a combination of waiting times for the two buses. A diagonal line from (0,0) to (5,5) divides the area in two - on one side the 5-minute bus comes first, on the other the eight-minute bus. From that the probability -

(5 min x 8 min - ½ x 5 min x 5 min) / (5 min x 8min) = 5.5 / 8 = 11/16

The second case could be done that way.. cutting some shape out of a 3 x 5 x 8 shoe box.. but it gets harder to visualize, so I did with a couple of ifs. The five-minute bus can only come first if it comes in three minutes. If the eight-minute bus also comes within three minutes, there is an equal chance for any of the three buses; if not, the equal chance is between the five-minute and the 3-minute bus. Thus,

1/3 x 3/5 x 3/8 + 1/2 x 3/5 x 5/8 = 18/240 + 45/240 = 63/240 = 21/80

Both of these concur with the answers of Forkedknight.