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If an event occurs an average of 0.6 times/year over a century, what is the probability that it occurs exactly once in a randomly selected year?

I was able to find p(occurs exactly once in a given year) for a 10-year period by calculating the probability of all 6 occurrences being in different years, of 4 in different years, of 3 in different years and so on, then multiplying each by the resultant probability of a randomly selected year having exactly one occurrence (6/10, 4/10, 3/10,...). However, it would be highly laborious to apply this method to a 100-year period.

How could this be done more efficiently?

pfroehlich2004
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    Of course this answer will depend on the distribution, but I'd say the Poisson Distribution was a natural place to start. In that case we'd get $p(1)=\frac {.6^1e^{-.6}}{1!}\sim .33$ – lulu Jun 25 '16 at 15:18
  • Hmm. A period of 5 years would be the smallest for which a mean of 0.6 gives the total number of occurrences as an integer. For 3 occurrences in 5 years, there is a total of 35 combinations of years in which they could happen. There are 10 combinations in which all 3 events occur in separate years, 20 in which 2 occur in the same year and 1 occurs in another year, and 5 in which all 3 occur in the same year. If the events occur in 3 separate years (p=2/7), then p(exactly 1 in randomly selected year)=3/5. If 2 events occur in the same year and 1 in another (p=4/7), then p(exactly 1)=1/5 – pfroehlich2004 Jun 27 '16 at 04:59
  • (2/7)(3/5)+(4/71/5)=0.286, which is quite a bit smaller than the prediction given by the Poisson Distribution. Repeating this for a period of 10 years with 6 events gives us an even lower value (p=0.257). Have I made an error in my calculations? – pfroehlich2004 Jun 27 '16 at 05:49
  • As I said, the answer will strongly depend on the distribution....the information you give does not determine the answer uniquely. I can make it effectively $0$ if you like (suppose the thing is binary, with $0$ occurrences in a given year with probability $.9999$ and $60000$ in a year with probability $.0001$. The Poisson process has the wonderful property that the probability of occurrence is equal in equal time intervals...I thought (and still think) that was the most natural reading of your conditions. But the answer will depend on the distribution. – lulu Jun 27 '16 at 10:04

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