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Math Help - Probability - Drivers and accidents

  1. #1
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    Question Probability - Drivers and accidents

    Another probability question, I've just started doing stats and probability, I've already done 23 questions now haha, this is number 42 in a longgggg list of examples im trying to master to form a masterbook so i can solve most problems . I've got as far as defining what I know etc, i.e. P(XA)=0.02 (probability that X has accident) any help would be cool

    An insurance company classifies drivers as class X, Y or Z. Experience indicates that the probability a class X driver has at least one accident in any one year is 0.02, while the corresponding probabilities for Y and Z are 0.04 and 0.1 respectively. The class of a new driver who applies to them for cover is not known, but from experience they have found that
    30% of applicants are class X, 60% are class Y , and 10% are class Z.

    (a) What the the probability of a new client having an accident in their first year?

    (b) If a new client has no accidents in their first year, what is the probability that they are a class X driver?
    (c) What is the probability of a new client having no accidents in their first 5 years?
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  2. #2
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    Quote Originally Posted by mitch_nufc View Post
    Another probability question, I've just started doing stats and probability, I've already done 23 questions now haha, this is number 42 in a longgggg list of examples im trying to master to form a masterbook so i can solve most problems . I've got as far as defining what I know etc, i.e. P(XA)=0.02 (probability that X has accident) any help would be cool


    An insurance company classifies drivers as class X, Y or Z. Experience indicates that the probability a class X driver has at least one accident in any one year is 0.02, while the corresponding probabilities for Y and Z are 0.04 and 0.1 respectively. The class of a new driver who applies to them for cover is not known, but from experience they have found that
    30% of applicants are class X, 60% are class Y , and 10% are class Z.

    (a) What the the probability of a new client having an accident in their first year?

    (b) If a new client has no accidents in their first year, what is the probability that they are a class X driver?

    (c) What is the probability of a new client having no accidents in their first 5 years?
    Pr(at least 1 accident | X) = 0.02

    Pr(at least 1 accident | Y) = 0.04

    Pr(at least 1 accident | Z) = 0.1

    Pr(X) = 0.3

    Pr(Y) = 0.6

    Pr(Z) = 0.1


    (a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

    (b) Calculate Pr(X | no accident).

    Note: Pr(no accident) = 1 - answer to (a) = ....
    Pr(X and no accident) = Pr(no accident | X) Pr(X) = ....

    (c) (1 - answer to a)^5, assuming each year is independent.
    Last edited by mr fantastic; February 7th 2009 at 12:53 PM. Reason: Fixed a careless slip-up
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  3. #3
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    I got 3.2 for part b), this obviously isnt right at all lol
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    Quote Originally Posted by mitch_nufc View Post
    I got 3.2 for part b), this obviously isnt right at all lol
    Post your calculations.
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  5. #5
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    P(no accident|X)=P(X & no accident)/P(X)

    =0.96/0.3

    as P(X & no accident)=P(X)P(no accident|X)

    Should I have got 0.288?
    Last edited by mr fantastic; March 2nd 2009 at 04:33 AM. Reason: Merged posts
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    Quote Originally Posted by mitch_nufc View Post
    Another probability question, I've just started doing stats and probability, I've already done 23 questions now haha, this is number 42 in a longgggg list of examples im trying to master to form a masterbook so i can solve most problems . I've got as far as defining what I know etc, i.e. P(XA)=0.02 (probability that X has accident) any help would be cool

    An insurance company classifies drivers as class X, Y or Z. Experience indicates that the probability a class X driver has at least one accident in any one year is 0.02, while the corresponding probabilities for Y and Z are 0.04 and 0.1 respectively. The class of a new driver who applies to them for cover is not known, but from experience they have found that
    30% of applicants are class X, 60% are class Y , and 10% are class Z.

    (a) What the the probability of a new client having an accident in their first year?

    (b) If a new client has no accidents in their first year, what is the probability that they are a class X driver?
    (c) What is the probability of a new client having no accidents in their first 5 years?
    Mitch,

    On (b), use
    P(\text{X} \; | \; \text{no accident}) =\frac{P(\text{X and no accident})}{P(\text{no accident})}

    On (c), use
    P(\text{no accident})^5.

    (I think Mr. F. slipped a little on that last one, if I understand the problem correctly.)
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    Quote Originally Posted by awkward View Post
    [snip]On (c), use
    P(\text{no accident})^5.

    (I think Mr. F. slipped a little on that last one, if I understand the problem correctly.)
    Yes, a careless slip-up. I've fixed it up and edited in what I meant.
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  8. #8
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    Could anyone show me how b) and c) are done with numbers put it I'm really tired and have a banging headache, all ive done all day is stats and bayesian stats and my head is mush thanks
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    Quote Originally Posted by mr fantastic View Post
    Pr(at least 1 accident | X) = 0.02

    Pr(at least 1 accident | Y) = 0.04

    Pr(at least 1 accident | Z) = 0.1

    Pr(X) = 0.3

    Pr(Y) = 0.6

    Pr(Z) = 0.1


    (a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

    (b) Calculate Pr(X | no accident).

    Note: Pr(no accident) = 1 - answer to (a) = ....
    Pr(X and no accident) = Pr(no accident | X) Pr(X) = Mr F adds: (0.98) (0.3).

    (c) (1 - answer to a)^5, assuming each year is independent.
    Quote Originally Posted by mitch_nufc View Post
    Could anyone show me how b) and c) are done with numbers put it I'm really tired and have a banging headache, all ive done all day is stats and bayesian stats and my head is mush thanks
    Advice: Take a break. Then either re-do the calculation using what's in this post or .....

    Post your calculation with all the details. The calculations you posted did not have detail to make it easy for me to see what you've done:

    Quote Originally Posted by mitch_nufc View Post
    P(no accident|X)=P(X & no accident)/P(X)

    =0.96/0.3

    as P(X & no accident)=P(X)P(no accident|X)
    Then, if you've made a mistake, it will be easy to find.

    Note that I have given you all the numbers that go into the formula that awkward provided and which it appears you're using. The rest is just arithmetic.
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  10. #10
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    Question

    (a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

    (b) Calculate Pr(X | no accident).

    Note: Pr(no accident) = 1 - answer to (a) = ....
    Pr(X and no accident) = Pr(no accident | X) Pr(X) =
    Mr F adds: (0.98) (0.3).

    (c) (1 - answer to a)^5, assuming each year is independent.

    so from part a) we get 0.04.
    therefore 1- a) = 1-0.04 = 0.96 not 0.98 yeah?
    so part b) = 0.96x0.3
    & part c) = (0.96)^5
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  11. #11
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    Quote Originally Posted by mitch_nufc View Post
    (a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

    (b) Calculate Pr(X | no accident).

    Note: Pr(no accident) = 1 - answer to (a) = ....
    Pr(X and no accident) = Pr(no accident | X) Pr(X) = Mr F adds: (0.98) (0.3).

    (c) (1 - answer to a)^5, assuming each year is independent.
    so from part a) we get 0.04.
    therefore 1- a) = 1-0.04 = 0.96 not 0.98 yeah? Mr F says: No.

    so part b) = 0.96x0.3
    & part c) = (0.96)^5
    Pr(no accident) is not the same as Pr(no accident | X).

    Pr(no accident | X) = 1 - Pr(at least 1 accident | X) = 1 - 0.02 = 0.98 like I said.
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  12. #12
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    FINALLY done this Q!

    Thanks to Mr. F and awkward for combined help.
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