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Math Help - probability of one-car accident?

  1. #1
    Yan
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    probability of one-car accident?




    The probability that a one-car accident is due to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one-car accident is incorrectly attributed to faulty brakes is 0.03. what is the probability that
    a) a one-car accident will be attributed to faulty brakes;
    b) a one-car accident attributed to faulty brakes was actually due to faulty brakes?
    Last edited by Yan; January 18th 2009 at 06:06 PM.
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  2. #2
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    Quote Originally Posted by Yan View Post

    The probability that a one-car accident is correctly attributed to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one-car accident is incorrectly attributed to faulty brakes is 0.03. what is the probability that
    a) a one-car accident will be attributed to faulty brakes;
    b) a one-car accident attributed to faulty brakes was actually due to faulty brakes?
    a) Will occur when either the accident is CORRECTLY attributed to faulty brakes, OR when the accident is INCORRECTLY attributed to faulty brakes. If an event occurs when either one event OR another event happens, then you simply add their probabilities. Hence 0.04+0.03.

    b) This will occur when two things happen simultaneously. The accident happened due to faulty brakes, AND the accident was attributed to faulty brakes. When we want the probabilities of two things happening simultaneously, we multiply their probabilities. Hence (0.04)(0.03).
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  3. #3
    Yan
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    Quote Originally Posted by Mush View Post
    a) Will occur when either the accident is CORRECTLY attributed to faulty brakes, OR when the accident is INCORRECTLY attributed to faulty brakes. If an event occurs when either one event OR another event happens, then you simply add their probabilities. Hence 0.04+0.03.

    b) This will occur when two things happen simultaneously. The accident happened due to faulty brakes, AND the accident was attributed to faulty brakes. When we want the probabilities of two things happening simultaneously, we multiply their probabilities. Hence (0.04)(0.03).
    it is not the right answer
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  4. #4
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    Quote Originally Posted by Yan View Post


    The probability that a one-car accident is due to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one-car accident is incorrectly attributed to faulty brakes is 0.03. what is the probability that
    a) a one-car accident will be attributed to faulty brakes;
    b) a one-car accident attributed to faulty brakes was actually due to faulty brakes?
    It helps to first write out the given probabilities:

    Pr(cause of accident is attributed to faulty brakes | brakes are faulty) = 0.82.

    Pr(cause of accident is attributed to faulty brakes | brakes are NOT faulty) = 0.03.

    Pr(accident due to faulty brakes) = 0.04.


    (a) Pr(cause of accident is attributed to faulty brakes) = (0.82)(0.04) + (0.03)(0.96) = ....


    (b) You need to calculate Pr(brakes are faulty | cause of accident is attributed to faulty brakes). Note that:

    Pr(brakes are faulty | cause of accident is attributed to faulty brakes) Pr(cause of accident is attributed to faulty brakes) = Pr(cause of accident is attributed to faulty brakes | brakes are faulty) Pr(brakes are faulty).

    Therefore:

    Pr(brakes are faulty | cause of accident is attributed to faulty brakes) (answer to part (a)) = (0.82) (0.04) ....
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  5. #5
    Yan
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    Quote Originally Posted by mr fantastic View Post
    It helps to first write out the given probabilities:

    Pr(cause of accident is attributed to faulty brakes | brakes are faulty) = 0.82.

    Pr(cause of accident is attributed to faulty brakes | brakes are NOT faulty) = 0.03.

    Pr(accident due to faulty brakes) = 0.04.


    (a) Pr(cause of accident is attributed to faulty brakes) = (0.82)(0.04) + (0.03)(0.96) = ....


    (b) You need to calculate Pr(brakes are faulty | cause of accident is attributed to faulty brakes). Note that:

    Pr(brakes are faulty | cause of accident is attributed to faulty brakes) Pr(cause of accident is attributed to faulty brakes) = Pr(cause of accident is attributed to faulty brakes | brakes are faulty) Pr(brakes are faulty).

    Therefore:

    Pr(brakes are faulty | cause of accident is attributed to faulty brakes) (answer to part (a)) = (0.82) (0.04) ....
    thanks for your help! I think my problem is confusing the problem.
    thank you so much
    BUT, I HAVE ONE MORE PROBLEM WITH IT, THE LAST PART
    Pr(brakes are faulty | cause of accident is attributed to faulty brakes) (answer to part (a)) = (0.82) (0.04)...
    IS IT SHOULD BE
    (0.82) (0.04)/(answer to part (a))?
    Last edited by Yan; January 19th 2009 at 11:23 AM.
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