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Math Help - Probability problem

  1. #1
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    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
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  2. #2
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    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: E(X) = 0.8 \times 0 + 0.11 \times 1 + 0.05 \times 2 + 0.04 \times 3
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  3. #3
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    Quote Originally Posted by nzmathman View Post
    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: 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.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    This gives the expected number of repairs in one year.
    Sorry, must have missed that "given year" part somehow
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