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My friend is taking a course on probabilities and come across a problem that has a solution we both have trouble understanding:

A single printer prints, on average, 22 jobs an hour. Let's use a simplification where no job takes more than 2 minutes to complete. Using these assumptions, calculate the upper limit for the probability that, when a job arrives to the printer, the printer is processing another job.

The solution was to use the Poisson distribution: 1 - ((22/30)^0)*e^(-22/30)/1! = 1 - e^(-22/30) = ~0.52

I'm having trouble believing this: wouldn't simply 22/30 make more sense? Even though I can see how enqueueing the printing jobs is Poisson distributed, how come that goes for the waiting times, too? Or have they messed up with the solutions?

Edit: after sending feedback the assistant said the problem was, indeed, ambiguous. Thanks for the help.

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I don't think the problem is right (though perhaps it's a reasonable approximation). It implicitly allows other jobs to overlap! –  Charles Oct 3 '10 at 23:01
    
The problem needs additional assumptions. For example, suppose jobs are submitted in bursts of 11 on the hour: then the probability is 10/11 that a job will be queued while another is printing. –  whuber Oct 4 '10 at 14:13

1 Answer 1

Update This was my first answer:

Your answer 22/30 ignores the fact that the job preceding yours may itself have been held up by a previous job.

As Charles and tsiki point out, this is wrong: it's the Poisson-derived answer that ignores this. So now I'll try to answer the question as posed:

By symmetry, the probability of a job having been started less than 2 minutes prior to yours is the same that a job will be started less than 2 minutes after yours. This is a Poisson distribution.

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Sorry, I'm not sure I follow... I thought the solution which utilizes Poisson distribution allows the waiting times to overlap each other (as Charles above has pointed out) and ignores the fact that the printing jobs might've been held up by previous jobs? –  tsiki Oct 4 '10 at 0:03

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