Question

(7) An actuary for an insurance company created the following probability table for a certain home insurance policy for a home in Drachtown, USA (which is a very depressing place). Event Payout to Policy Holder. Probability No claim $0 0.8855 Minor claim $4,500 0.0905 Medium claim $16,800 0.0142 Major Claim $70,500 0.0098 (a) Presuming years are independent, determine the probability that over the course of 12 years, the policy holder is paid some money at least once. Show calculation. (b) Determine the expected payout each year. Show calculation. (c) Find the probability of a medium claim six years in a row. Show calculation.

Answer 1

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a) let us define Y= 1 if there is a claim and Y=0 if there is no claim in a year. Then X = SY follows binomial distribution with parameters n=12 and p= 1-0.8855 = 0.1145, The pmf of X is P(X= =)=(*)*(1-2)** = P(X=»)= (142)*1-2)** Now the required probability is P(X>0) = 1 - P(X=0) = 1-1 12 0.1145° (1-0.1145) = 0.7676 10) b) The expected pay-out in each year is (0*0.8845) +(4500*0.0905) +(16800*0.0142) +(70500*0.0098) = $1336.71 c) The probability of medium claim in six years a row is 0.0142 = 8.198418e-12