There are three different periods of unequal lengths where one specific event is happening such that it follows Poisson distributions (e.g. meteorite falls):

  • in the 1st period (100 hours) the average interval for event is 500 hours
  • in the 2nd period ( 80 hours) the average interval for event is 700 hours
  • in the 3rd period (130 hours) the average interval for event is 300 hours


  • 1st period lambda: (1/500)*100 = 0.200
  • 2nd period lambda: (1/700)*80 = 0.114
  • 3rd period lambda: (1/300)*130 = 0.433

Probability to see at least 1 event (complement to P(0 events):

  • 1st period prob.: 1 - (e^-0.200 * (0.200^0/0!)) = 0.181
  • 2nd period prob.: 1 - (e^-0.114 * (0.114^0/0!)) = 0.108
  • 3rd period prob.: 1 - (e^-0.433 * (0.433^0/0!)) = 0.352

Now, what is the probability that at least one event happened over the three periods? How do I add the probabilities (or the numbers before I calculated probabilities) together?

  • 2
    $\begingroup$ Apply the assumption of independence (of events in non-overlapping intervals) to compute the chance that no event happened and from that you will easily obtain the answer. $\endgroup$
    – whuber
    Commented Oct 16, 2019 at 19:57

1 Answer 1


The probability that one or more events happens in any of the three periods is 1 - the probability that 0 events happen in the three periods. Assuming that these periods are independent we can simply multiply the probabilities together and then do 1- that probability.

So in your case it is: Probability of no events 1st period: 0.819 2nd period: 0.892 3rd period: 0.648

So we need to multiply them together to get the probability that 0 events happen in all three periods: 0.473

Now to get the probability of at least one event happening over these three periods we need to do 1-0.473 to get a probability of ~0.526

For a thorough explanation behind combining multiple independent probabilities please see here: https://www.mathgoodies.com/lessons/vol6/independent_events


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