Jim has a 5-year-old car in reasonably good condition. He wants to take out a $3
ID: 3323477 • Letter: J
Question
Jim has a 5-year-old car in reasonably good condition. He wants to take out a $30,000 term (that is, accident benefit) car insurance policy until the car is 10 years old. Assume that the probability of a car having an accident in the year in which it is x years old is as follows:
Jim has a 5-year-old car in reasonably good condition. He wants to take out a $30,000 term (that is, accident benefit) car insurance policy until the car is 10 years old. Assume that the probability of a car having an accident in the year in which it is x years old is as follows 7 x age 0.01191 0.01292 0.01396 0.01503 0.01613 P(accident) Jim is applying to a car insurance company for his car insurance policy. The expected loss to the car insurance company for the 5th, 6th, 7th, 8th, and 9th years would be S357.30, S387.60, S418.80, S450.90 or $483.90 respectively. What would be the total expected loss to the car insurance company over the years 5 through 9? Round your answer to the nearest dollar.Explanation / Answer
Probability of the paying out the loss in the 1st year of coverage is .01191.
The probability of paying out the loss in the 2nd year of coverage is (.98809)*(.01292)= .012766. The (.98809) term is from the probability the car was accident-free in the 1st year (i.e. 1-.01191=.98809).
Using the same logic, the the probability paying out the loss in the 3rd year is (.98809)*(.98708)*(.01396)= .0136155.
Paying out loss in 4th year ---> (.98809)*(.98708)*(.98604)*(.01503) = .0144545
Paying out loss in 5th year ---> (.98809)*(.98708)*(.98604)*(.98497)*(.01... = .015279
Expect loss is therefore 30,000 (.01191 + .012766 + .0136155 + .0144545 + .015279) = 2040.75
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