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Probability that the driver is from G3

  1. Jan 20, 2009 #1
    There are three groups of drivers in a city: G1, G2, and G3.

    G1 make up 30% of drivers, G2 make up 50% of drivers, and G3 make up the remaining 20% of drivers.

    P(at least one accident for G1)=0.1 = P(A1)
    P(at least one accident for G2)=0.3 = P(A2)
    P(at least one accident for G3)=0.5 = P(A3)

    If we randomly select a driver out of the groups and the driver has no accident, what is probability that the driver is from G3?

    I tried [tex]P(G3 | (A1^{c} \cap A2^{c} \cap A3^{c}))[/tex]. Is this is the right method of solving such a question? The probability I got from this happened to be zero which sort of confirms that it's incorrect. Any advice will be great.

    Thanks a lot.
  2. jcsd
  3. Jan 21, 2009 #2
    Re: Probability

    I like the brute force way since I don't know the notation. Assuming one hundred drivers in proportions given, 30 G1 drivers, etc. 1- prob of one accident is prob of no accidents, right ? So there are 27 no accident G1 drivers. Add up all the no accident drivers for G2 and G3, should be 72. What proportion of these 72 are G3 ? Well, there are 10 no accident G3 drivers, so 10/72 is probability.
  4. Jan 22, 2009 #3
    Re: Probability

    It would be nice if the answer could be put in a form more in line with advanced probability thinking. So...?

    For conditional probability: [tex]P(A\mid B)=\frac{P(A and B)}{P(B)}[/tex]

    So in this case A represents G3 and B represents no accident.

    [tex]\frac{20*.5}{30*.9+50*.7+20*.5} = 10/27 [/tex]

    I would not say that this "improves" the answer whatsoever, or even whether it applies.
    Last edited: Jan 22, 2009
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