Probability problem: waiting for the bus

Hi all,

I've got a problem that's giving me some trouble:

A boy goes to the bus stop at some time between 8am and 9am, each moment being equally likely. The bus will come some time between 7:30am and 8:30am, also equally likely. The bus will wait up to 15 minutes for the kid. The kid will wait up to 30 minutes for the bus. What is the chance the kid makes it onto the bus?


What's tripping me up is that the boy and the bus wait for different amounts of time and their arrive times aren't spread out over the same hour long time interval. If they both arrived between 8 and 9 and both waited 15 minutes then I think I could solve it simply by plotting it out on in a square on the xy-plane and finding the area in a certain region of the square. But I'm having trouble finding a way to do it with this setup.

Thanks in advance for any pointers!

Stochastic_Jimmy said:

A boy goes to the bus stop at some time between 8am and 9am, each moment being equally likely. The bus will come some time between 7:30am and 8:30am, also equally likely. The bus will wait up to 15 minutes for the kid. The kid will wait up to 30 minutes for the bus. What is the chance the kid makes it onto the bus?

Click to expand...

How about using trapezoidal distributions instead of square?

For instance, Bus: from 8:00 to 8:15, the probability of seeing the bus increases from zero up to a value that is constant till 9:00, then drops linearly to 0 at 9:15. Normalize the distribution so that the area is one. [Note that the independent variable has units of "hours," so the pdf has units of "per hour."]

Boy: similar

How then can you find the probability that both are there at the same time?

DrPhil said:

How about using trapezoidal distributions instead of square?

For instance, Bus: from 8:00 to 8:15, the probability of seeing the bus increases from zero up to a value that is constant till 9:00, then drops linearly to 0 at 9:15. Normalize the distribution so that the area is one. [Note that the independent variable has units of "hours," so the pdf has units of "per hour."]

Boy: similar

How then can you find the probability that both are there at the same time?

Click to expand...

Thanks Dr. Phil. I've give it a try.

Out of curiosity, can anyone think of any other approaches? I'm pretty sure trapezoidal distributions were not assumed background knowledge in the context in which this question was posed.

Thanks a bunch!

Stochastic_Jimmy said:

Thanks Dr. Phil. I've give it a try.

Out of curiosity, can anyone think of any other approaches? I'm pretty sure trapezoidal distributions were not assumed background knowledge in the context in which this question was posed.

Thanks a bunch!

Click to expand...

The trapezoid is the result of convoluting two square distributions, one an hour wide and the other 1/4 hour (for the bus). If you imagine passing the short one across the wide one, and asking what is the area of the intersection as a function of the time offset, you should see how that comes out to be a trapezoid. At any specific time, the probability of seeing a bus is equal the the probability of a bus having arrived at the stop any time in the preceding 15 minutes.

Stochastic_Jimmy said:

Thanks Dr. Phil. I've give it a try.

Out of curiosity, can anyone think of any other approaches? I'm pretty sure trapezoidal distributions were not assumed background knowledge in the context in which this question was posed.

Thanks a bunch!

Click to expand...

The way I'd probably go about it, is by listing the various times which the bus may show up against the various times that the boy may show up, in 15 minute blocks, seeing as this is the minimum denomination of time in the problem.

The boy and the bus therefore have 4 various 15-minute blocks of time with which they may show up. This presents 16 combinations. You can then analyse each of the 16 combinations in a truth-table, to determine whether the probability of the boy getting on the bus for that specific combination is 0%, 50%, or 100%.

For example:

Boy showing up 7:45 - 8, and bus showing up 8:30 - 8:45 represents a 0%

Boy showing up 7:45 - 8, and bus showing up 8:15 - 8:30 represents a 50% because the 30 minute wait time can be interpreted as the boy leaving the bus stop due to the bus not arriving between 8:15 - 8:30. Since the probability of all times is equally possible, this means that it's an equal chance of either the bus coming whilst the boy is waiting, or after the boy has stopped waiting.

Boy showing up 7:45 - 8, and bus showing up 8 - 8:15 represents a 100%