Looks to me like a simple problem calculating the expected value. The expected number of repairs for the given freezer, E(X) = sum of: each outcome times its probability.
In this case:
A fast food restaurant just leased a new freezer for 3 years. The service contract offers unlimited repairs for a fee of $125 a year plus a $35 service charge for each repair needed. The restauran't research suggested that during a given year 80% of these freezers need no repair, 11% needed to be serviced once, 5% twice, 4% three times and none required more than three repairs.
How many times should the retaurant expect to have to get this freezer repaired over the three-year term of the leases?
Is this .20 p(needing repair) x 3 years? or .20*.20*.20
This gives the expected number of repairs in one year.
A simple but tedious approach would be to draw a tree diagram and use it to calculate the probability of zero, one, two and three repairs over a three year period. The calculate the expected value of the number of repairs over a three year period in the usual way.