The way to present the events happening in each 5-year interval is to write the 5-letter string

like this, for example (N Y N N Y), which represents "No flood" in the years 1,3,4 and "Yes flood" in the years 2,5.



Then the events having flood in 3 years in a row of 5 years are these three events

    (Y Y Y N N),  (N Y Y Y N)  and  (N N Y Y Y)

The corresponding probabilities for each of these three events are   =  = 0.02304.



Next, the events having flood in 4 years in a row of 5 years are these two events

    (Y Y Y Y N)  and  (N Y Y Y Y).

The corresponding probabilities for each of these two events are   =  = 0.01536.



Finally, the event having flood in 5 years in a row of 5 years is this unique event

    (Y Y Y Y Y).

The corresponding probability for this unique event is   = 0.01024.



The ANSWER  is the sum of these probabilities taken with their multiplicities  


    P = 3*0.02304 + 2*0.01536 + 0.01024 = 0.11008.    A N S W E R