<Dd> P (k = 2 overflow floods in 100 years) = 1 2 e − 1 2! = e − 1 2 ≈ 0.184 (\ displaystyle P (k = 2 (\ text (overflow floods in 100 years))) = (\ frac (1 ^ (2) e ^ (- 1)) (2!)) = (\ frac (e ^ (- 1)) (2)) \ approx 0.184) </Dd> <P> The table below gives the probability for 0 to 6 overflow floods in a 100 - year period . </P> <Table> <Tr> <Th> k </Th> <Th> P (k overflow floods in 100 years) </Th> </Tr> <Tr> <Td> 0 </Td> <Td> 0.368 </Td> </Tr> <Tr> <Td> </Td> <Td> 0.368 </Td> </Tr> <Tr> <Td> </Td> <Td> 0.184 </Td> </Tr> <Tr> <Td> </Td> <Td> 0.061 </Td> </Tr> <Tr> <Td> </Td> <Td> 0.015 </Td> </Tr> <Tr> <Td> 5 </Td> <Td> 0.003 </Td> </Tr> <Tr> <Td> 6 </Td> <Td> 0.0005 </Td> </Tr> </Table> <Tr> <Th> k </Th> <Th> P (k overflow floods in 100 years) </Th> </Tr>

Distribution of the sum of poisson random variables