Adding probabilities of different Poisson distributions

Question

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

Lambda:

• 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?

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