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Probability and insurance

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  1. Jul 27, 2005 #1
    Hello all

    I found this question rather interesting and very confusing, check it out

    an insurance company classifies its customers as either high risk 5% of people, medium risk 25% of people, and low risk 70% of people. now the probability of a high risk customer making a claim in the first year is 75%, medium risk 45% and low risk is 5%, now if the companies prcocedure is to place new customers in high risk,and if they dont make a claim in first year then they will be placed into medium risk in the second year and if they dont make a claim in the second year then they will be placed into low risk.

    firstly what is the probability of a customer going through this procedure if
    actually he is a low risk customer.
    secondly what is the probability of a customer going through this procedure if actually he is a high risk customer.
    lastly would such a procedure be fair on customers?

    I was only able to do the first bit, dont know if I am correct, please correct me if im wrong
    P(Procedure|low risk)=P(does not make claim&high risk)+
    P(does not make claim&medium risk)=0.05*0.25+0.25*0.55=0.15

    steven
     
  2. jcsd
  3. Jul 27, 2005 #2

    EnumaElish

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    I have only an imperfect sense of what it means for "a customer going through this procedure." Because, by definition, all new customers go through it, don't they? I am guessing that what it means is "a customer going through all of it and coming at the end as being classified as a low-risk customer." Is that the right understanding?

    And, about fairness: from whose point of view? It may be great for low-risks but costly for high-risks, or vice versa. Do you mean "someone who doesn't know his/her risk type," e.g. "the average customer" (an imaginary construct if there ever was one)?
     
  4. Jul 28, 2005 #3
    well here is a more clearer version of the problem this might help (some numbers have changed by the way)

    a motor vehicle insurance company divides policyholders into 3 catagories: high, medium and low risk. the high risk group contains 10% of people, and they each have a 35% chance of making a claim during a particular year. the medium risk contains 20% of people, and they each have a 15% of chancw of making a claim during a yaer. the low risk group contains the remaining 70% of people, and they each have a 5% chance of making a claim during a year. the company's general procedure is to put a new policy holder into the risk group in the first year. if the person doesnt make a claim, they move into the medium risk group in the second year, and then into the low risk group in the third year.what is the probability that this will happen if the person is actually a low risk driver? what if they are actually a high risk driver? does the companys procedure seem to be fair?

    well there it is, well out of all my knowledge about probability i cant answer this problem, i got a feeling there is a problem with the question, anyway any suggestions would be helpful

    steven
     
  5. Jul 29, 2005 #4
    hello all

    would anybody know what sub-topic this might relate to? I have tried looking for a similar problem but that didnt help, i have looked at conditional probability, but i cant see how it relates to it?

    steven
     
  6. Jul 30, 2005 #5

    EnumaElish

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    P1[high|low] = 1 (by assumption)
    P2[med|low] = 0.95
    P3[low|low] = 0.95
    Therefore
    P[low|low] = 0.952
     
  7. Jul 30, 2005 #6
    hello there

    so does this mean that

    P[of being placed into low risk after 2 years|that the customer is actually low risk]=P1[high|low]*P2[med|low]*P3[low|low]
    but how does one obtain these probabilities:

    P1[high|low] = 1 (by assumption)
    P2[med|low] = 0.95
    P3[low|low] = 0.95
    Therefore
    P[low|low] = 0.952

    from the information given all I can obtain is the following

    P[H]=0.1
    P[M]=0.2
    P[L]=0.7

    P[C|H]=0.35
    P[Cnot|H]=0.65

    P[C|M]=0.15
    P[Cnot|M]=0.85

    P[C|L]=0.05
    P[Cnot|L]=0.95

    P[CnH]=0.035
    P[CnotnH]=0.065

    P[CnM]=0.03
    P[CnotnM]=0.17

    P[CnL]=0.035
    P[CnotnL]=0.665

    P[C]=0.1
    P[Cnot]=0.9

    P[H|C]=0.35
    P[H|Cnot]=0.072222222

    P[M|C]=0.3
    P[M|Cnot]=0.188888888

    P[L|C]=0.35
    P[L|Cnot]=0.738888888

    and how does one draw a tree diagram for such problem, I have attached a diagram of my tree diagram check it out and update me if I have made any mistakes

    steven
     

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  8. Jul 31, 2005 #7

    EnumaElish

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    See, the problem asks "what is the probability of a low risk person going through the procedure?" In this question the person's type is a given. One does not need 0.1, 0.2, 0.7, because the probability of being type X for a type X person is 1.00. Heretofore:

    I know I am low risk (given in the problem). In the 1st year being placed in the H category is 1.00 (because the problem says so). At the end of my 1st year with 95% chance I will sail into the M category (with 0.05 prob I will stay on as H). In my 2nd year I have 0.95 prob to begin my 3rd year in category L. (With prob 0.05 I will stay in the M category.)

    Suppose I filed a claim in the 1st year and stayed as an H for a 2nd year. If in the second year I do not file a claim (as will be the case with 0.95 prob) then I will end up in the M category at the end of 2 years.

    So, GIVEN THAT I AM A LOW-RISK PERSON:
    Being placed in the L category at the end of 2 years is 0.95 squared.
    Being placed in the M category is 0.95 * 0.05 + 0.05 * 0.95.
    Being placed in the H category is 0.05 * 0.05.
    They add up to 1.00, as they should.

    You can derive a similar reasoning for a high risk type.

    Your tree in the attachment is for a person who does not know what type he or she is; because it starts with 3 branches with attached probs 0.1, 0.2, 0.7. If someone knows his or her risk type then they already know which of these 3 branches he or she is on, and does not need the other two branches. This is like giving you a probability problem that starts as "there is a 0.1 prob that an insurance enrollee's name is Charles, a 0.2 prob that it is William, and a 0.7 prob that it is Steven. All enrollees start out in the H category. For someone named Steven, there is a 0.95 chance that he will file a claim in his first year, etc." and then goes on to ask "what is the probability of ending up in the L category for someone named Steven?" Well, if you know your name is Steven, then you don't really need the 0.7 (and the 0.2 and the 0.1), do you? For you, being named Steven is a given, so it has a probability of 1.00.

    On the other hand if the problem had asked "for a random enrollee who may be named Charles, William or Steven with respective probs 0.1, 0.2, and 0.7, what is the probability of ...?" then you do need the 0.1, the 0.2, and the 0.7, because in this question, "being named Steven" is not a given, it is probabilistic.
     
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