Probability Question? Please Help.
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Probability Question? Please Help.

[From: ] [author: ] [Date: 12-05-17] [Hit: ]
P[1≤x≤3] = e^-13(13 + 13^2/2! + 13^3/3!) = .001048 -3/13.......
In a certain population an average of 13 new cases of esophageal cancer are diagnosed each year. If the annual incidence of esophageal cancer follows a Poisson distribution, find the probability that in a given year the number of newly diagnosed cases of esophageal cancer will be between 1 and 3 inclusive.

I am interested in how the question is done, not just the answer, I want to learn :) thank you in advance.

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the poisson formula is P[x] = e^-m *m^x/x! where m is the mean

P[1≤x≤3] = e^-13(13 + 13^2/2! + 13^3/3!) = .001048 <--------

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3/13.
13 is the number of possible outcomes
3 is the number of outcomes
3 outcomes(3) out of(/) 13 possible(13)
1
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