1. ## Probability problem

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

2. 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: $\displaystyle E(X) = 0.8 \times 0 + 0.11 \times 1 + 0.05 \times 2 + 0.04 \times 3$

3. Originally Posted by nzmathman
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: $\displaystyle E(X) = 0.8 \times 0 + 0.11 \times 1 + 0.05 \times 2 + 0.04 \times 3$
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.

4. Originally Posted by mr fantastic
This gives the expected number of repairs in one year.
Sorry, must have missed that "given year" part somehow