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In order t0 detenmine insurance premiums, suppose that car insurance company classifies policyholders into one of four classes: excellent risks; good risks, average...

Question

In order t0 detenmine insurance premiums, suppose that car insurance company classifies policyholders into one of four classes: excellent risks; good risks, average risks, and bad risks. Their records indicate that the probability that an excellent risk individual will be involved in car accident Ovcr one-year span is 0.02_ The sanie probability for good, average; and bad risk individuals are 0.05,0.14, and 0.32 ,respectively. Of its policy holders. 8"4 are classilied uS excellent risks; !6

In order t0 detenmine insurance premiums, suppose that car insurance company classifies policyholders into one of four classes: excellent risks; good risks, average risks, and bad risks. Their records indicate that the probability that an excellent risk individual will be involved in car accident Ovcr one-year span is 0.02_ The sanie probability for good, average; and bad risk individuals are 0.05,0.14, and 0.32 ,respectively. Of its policy holders. 8"4 are classilied uS excellent risks; !6% are classified as good risks; and 62% are classified as average Tisks: You maY aSSUIWe that Car accudents Are independent events. (A) Ovet e @iven OcYCat petiod; whut Ule puobbility Uau all policyholders will be involved in Cur aceident? Deline all GCIs epliculy Given that a policyhokler Was nt involved in cur aecident duting 2010, what is the probability the individual willbe classulied a5 (iVan execllent risk"? (I) Pood risk? (ITI) an average risk? (iv) a bad risk?



Answers

Car insurance companies categorize drivers as high risk, medium risk and low risk. (For your information only: Teens and seniors are considered high risk!) $20 \%$ of the drivers insured by 'First Insurance'are high risk, and $50 \%$ are medium risk driver. The company's actuaries estimate the chance that each class of driver will have at least one accident in the coming 12 months as follows: High risk $6 \%$ medium $3 \%$ and low $1 \%$ a) Find the probability that a randomly chosen driver is a high-risk driver that will have an accident in a 12 -month period. b) Find the probability that a randomly chosen driver insured by this company will have an accident in the next 12 months. c) A customer has a claim for an accident. What is the probability that he/she is a high-risk driver?

This problem. We're basically just given a state space with four states, and we're also given all the single step transition probabilities from all the states to all of the other states. That's all laid out in the question. And then, in part eighth, if we let x of n denote the state at n at stage end were asked to construct the one step transition matrix where this Markov chain were given in the question that the probability of going from state one to state one is 0.9. Probability of going from state one to state to is zero point 07 from state, one to state three is 0.2 and from state, one to state for his 0.1 So it may be easiest to you actually answered this question by simply filling in the Matrix as we go along. So P 11 goes into entry 11 in the Matrix. We have 0.9, 0.7 0.2 and 0.1 and it's always good to check that these add up to 1.1 It's always good to check that these add up to one and they do indeed add up to one. Every row should add up to one because they're the probabilities of going from the state represented by that road number. So in this case, state number one to all of the other possible states. So we know that it has to go from one state to one of the other possible states in the state space. We're also told that going from state 2 to 1 happens with probability of 0.8 and from going from 2 to 3 with probability, 0.15 and 2 to 4 is 0.5 and we can see that these probabilities already add up to one. So therefore, the probability of going from state to to to is zero next were given. The transition probability from ST 3 to 2 is 0.7 and 3 to 4 is also 0.7, or rather 0.3. So those add up to one. So therefore, these transition probabilities must be zero and were also given the transition probability from state four to state for is 0.4. Otherwise, you go from state four to state three, so that means that must be 0.6. So this is the single step transition matrix for this Markov chain than for B Were told that if you start in category one, what is the probability that you will be in category two five transitions later? So a good first step for that is to find the the fifth power of the single step transition matrix and using software that could be calculated as comes out to you the following. And so here is the five step transition matrix. And so the probability of going from state one to state to in five transitions is 0.918 now for C were asked if you are currently in state for what is the probability of being in state one in K years for K equals 1234 five and six. So that is the case, that probability of going from state four to state one. So for K equals one, we already have thes single step transition matrix. And for so going from state for two State one is zero in one transition, and we also have the five step transition matrix already so probability of going in five steps from state 4 to 1 is equal to you. 0.561 So we also need probabilities of doing it in two steps and in three, four steps and six steps. So I will go ahead and calculate these multi step transition matrices and I won't provide the full matrix for us. I would just take the the entry of four comma one, and that is our answer for each of these, so I'll provide that. And so it turns out that in two steps, the probability is still zero in three steps. The probability is 0.336 and four steps. We have zero point 437 and in six steps. That probability is 0.6 to 4 and last for party were asked, what is the probability of starting in state one, but then remaining in that state for the next five transitions? So when we calculated the five step transition transition matrix here, the probability of being in state one after five transitions is 0.813 but that also includes, so that includes the path where you just stay in state one. But it also includes all the other paths where you move out of state one into into eggs, for example, estate to but then moved back to ST one by the end of the fifth transition. So the actual probability for staying in ST one the entire time will be something less than 0.813 So we can't use the five step transition major export this instead. What we do is we take the single step transition probability, which is 0.9, and raise that to the power of five. And that comes out to you 0.59 So the probability of remaining in ST 135 transitions is 0.59

Here wants us to find the probability that a driver well of an accident during this year. So we know that, you know, to find this, we essentially need to first of all, add up, um, the probabilities of each different risk. So we know that we have our low risk insurance, medium risk insured and high risk injured and also gives us the one year period of accidents that are actually given. So with our first of all, with our lowest, um, drivers, we know that 60% are gonna be insured. Medium rooster is gonna be 30% insurance and hire service could be 10% that are injured within these accident rates are going to be as follows. So with a low amorous 1% medium risk your 5% and high risk 9%. So it us after this, we basically need to add up the probabilities so low risk plus medium risk. Plus, iris is going to give us the total accidents per year in this particular case. So with lowest, we know that it's gonna be 6/10 or 60% multiplied by is your point 1/10 or 1% of this particular case and we just and this to 30 over 100 or essentially 3/10 if every right that, um and we multiply this by 5% which is going to 0.5 out of 10. And we add, of course, one out of 10 multiplied by 0.9 out of 10. And that's going to give us a total of 0.6 over 100 if we added up, um, plus 1.5 over 100 and then that she have a seer 0.9 over 100 I should be 100 rather, and that should give us a total of three over 100 or, in other words, 3% risk or probability that the driver will have an accident during that particular year.

So we can see the table here. That is the age in the prophet. Oh, so the probability So, Adri, How 21 22 23 24. And if I, when you call a lot driven 26. So for the fourth year, do some one by 21 Rick Crawford. Well, bury my notes of 100 Sultan's plus 250 that ihsaa money that person paid for the first year. And also for the second year, we need to plus another 220 again, not there to hang greed and 30 year olds. You rode your role the other one so that it's a fright. And if the person rather than by for those years the profit will be to Hendry and comes well, 300 for a coupon fraud and the probability here we got all those probability. And for this one, the little P here thus equals one miners The rope on zero to your old 183 rhinos 0.186 rhinos point 00189 Miners 0.1 night one 193 Reach Yukos who point Sinai zero for out eight. And those offered exactly. While we know that equals who each profit hums each probability off it. So we can God, uh, Miyu here, just you call. So 303 0.35

So we see that an insurance company is offering four different deductible levels, non low medium and high. Um So the first thing we want to do is determine the probability that the individual has a medium audit deductible and a high homeowner's deductible. So given the different values we can um we have to consider them having probability from having both. Um So in this case, based on the data we're going to have 0.10 which means that 10% probably lives. That'll be the case. Now we want to know the probability that the individual has low and a low auto deductible and that's going to be 18% 4.18 And then for the low homeowner's deductible, that'll be 0.19 we can keep doing this. Um And we'll see that when we're doing the and statements we have to multiply them in or statements we will add.


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