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Poisson distributions for probability with less than

purplesoup
purplesoup used Ask the Experts™
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I understand how to work out probabilities where the number is greater than a given value, now I'm trying to work out how to get the probability of something less than a given value happening.

The simple answer is to just work out the greater than value, and subtract that from 1 to get the less than value, however this is missing the "equal to" part.

If I just work out "what is the probability that a given event will happen less than 3 times" I could work out the probability that it will happen more than 3 times, and subtract that value from 1. However doing that I will calculate the probability of the event happening less than or equal to three times.

How can I calculate an event happening just less than (not less than and equal to) - for example less than three times?

Example: cars crossing a junction during daylight hours may be modelled as a Poisson process with the rate of four per hour. What is the probability that the gap between two successive cars will be more than 20 minutes but less than 30 minutes?
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There are usually several ways to solve a problem and calculate an answer.
Figuring out which one to use, which one is easiest, is a matter of experience.

Look at the Birthday Problem for example:

     http://en.wikipedia.org/wiki/Birthday_problem

It is definitely easier to find the probability of 0 matches, than to find the probability of any number of matches.
>>  How can I calculate an event happening just less than (not less than and equal to)
      - for example less than three times?

Depends on the details of course, but there are at least two approaches:

    Find the probability for n = 3 or more and subtract it from 1.

    Find the probabilities for n=1, n=1, and n=2 and add them up.
also find the probability of = and subtract it from the probability of =<
what's important with this question, is that poisson is closely related to the exponential

the probability of the time to an event is given by the exponential distributions Cumulative distribution function

P(X <= x) = 1 - e^(- lamda x)

P(X > x) = 1 - P(X <= x)  = 1 - (1 - e^(- lamda x)) = e^(- lamda x)

P(X >20) = e^(- lamda 20)

P(X >30) = e^(- lamda 30)


P(between 20 and 30) = P(X >20)  - P(X >30)  = e^(- lamda * 20) - e^(- lamda * 30)

because time is continuous, you don't need to make a distinction between 'greater than 20 minutes' and 'greater than or equal to 20 minutes', becuase the difference in probability is infinitesimal (literally zero in fact) - due to the effects of limits

one last thing, you rate is 4 per hour, so your lamda for minutes is 4/60 = 1/5
but you could work with lamda = 4 and time in hours, e.g (1/3) and (1/2) of an hour - with the calculations becoming the same
purplesoupProgrammer

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Commented:
Thanks for your help.