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Math Help - please help...

  1. #1
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    please help...

    An automobile insurance company classifies drivers as Class A(good risks), class B(medium risks), and class C(poor risks). They believe that class A risks constitute 30% of the drivers who apply to them for insurance, class B 50%, and class C(20%). The probability that a class A driver will have one or more accidents in any 12 month period is .01, for class B is .03, and for class C is .10. The company sells Mr. Jones an insurance policy and within a year he has an accident. What is the probability that Mr. Jones is a class B risk?
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  2. #2
    Eater of Worlds
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    Are you familiar with Bayes' theorem?.

    Let's let D=portion of drivers from the respective classes

    P(B|D)=\frac{P(D|B)P(B)}{P(D|B)+P(D|A)P(A)+P(D|C)P  (C)}

    \frac{(.5)(.03)}{(.5)(.03)+(.30)(.01)+(.20)(.10)}=  .395

    The probability the driver is class B is about 39.5%.

    Check me out here. It's easy to get mixed up. It's been awhile since I used Bayes' theorem.
    Last edited by galactus; August 7th 2006 at 04:53 AM.
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  3. #3
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    Hello, bobby77!

    Galactus' answer is absolutely correct . . .


    An automobile insurance company classifies drivers as:
    . . Class A (good risks), class B (medium risks), and class C (poor risks).
    They found that class A constitutes 30% of their drivers, class B 50%, and class C 20%.
    The probability that a driver will have one or more accidents in any 12-month period is:
    . . 0.01 for class A, 0.03 for class B, and 0.10 for class C.

    The company sells Mr. Jones an insurance policy and within a year he has an accident.
    What is the probability that Mr. Jones is a class B risk?

    I checked the answer with a baby-talk approach.
    . . (You can use this when you're desperate.)

    Suppose the insurance company has exactly 1000 customers.

    Then 30% (300) are class A, 50% (500) are class B, and 20% (200) are class C.


    Of the 300 class A drivers, 1% will have an accident . . . 3

    Of the 500 class B drivers, 3% will have an accident . . . 15

    Of the 200 class C drivers, 10% will have an accident . . . 20

    Hence,  3 + 15 + 20 = 38 drivers will have accidents.


    Mr. Jones had an accident, so he is one of those 38 drivers
    . . . and 15 of them are class B drivers.

    Therefore, the probability that Mr. Jones is class B is:
    . . \frac{15}{38} \:=\:0.394736842 \:\approx\:39.5\%
    . . . ta-DAA!

    Nice work, Cody!

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  4. #4
    Eater of Worlds
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    Thanks for the confirmation, Soroban. I am glad I remembered Bayes' theorem.
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