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Math Help - Insurance Company accident probability

  1. #1
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    Insurance Company accident probability

    I'm a little unshure on how to solve this problem please help.

    The clayton Automobile Insurance Company only insures drivers that are 18 years of age. It has been determined that the probability that an 18 year old driver has an accident in a one year span is 24.65%. If the company insures twenty 18 year old drivers for a one year contract, determine the following.

    a) What is the probability that none of the drivers have an accident during their contract?

    b) What is the probability that at least 2 drivers have an accident during their contract?
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  2. #2
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    I think i figured this one out

    A)
    P(x=0)= 20C0(.2465)^0(.7535)^20
    P(x=0)= .00348

    B)
    P(x at least 2)= 1 - P(0) + P(1)
    1 - (20C0(.2465)^0(.7535)^20 + 20C1(.2465)^1(.7535)^19
    1- (.00348 + .0228) = .974
    P(x at least2) = .974

    Can some verify that these answers are correct please?

    We are both reviewing for a final exam and we need to know that we are prepared.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by hello there View Post
    I'm a little unshure on how to solve this problem please help.

    The clayton Automobile Insurance Company only insures drivers that are 18 years of age. It has been determined that the probability that an 18 year old driver has an accident in a one year span is 24.65%. If the company insures twenty 18 year old drivers for a one year contract, determine the following.

    a) What is the probability that none of the drivers have an accident during their contract?
    This one we can do straight off. All twenty must have no accident, so

    p=(1-0.2465)^20 ~= 0.003481

    But we may as well observe that the number of drivers who do have
    accidents has a binomial distrition with single case probability 0.2465,
    that is P(N drivers with accidents) ~ B(20, 0.2465).

    Then prob none have accidents is:

    b(0; 20, 0.2465) = 20!/[0! (20-0)!] 0.2465^0 (1-0.2465)^{20-0}

    ............................. = (1-0.2465)^20 ~= 0.003481

    b) What is the probability that at least 2 drivers have an accident during their contract?
    The probability that at least 2 drivers have an accident is the probability
    that 2 or more drivers have accidents which is 1 minus the probability that
    onle 1 or 0 drivers have accidents, so here:

    p(n>=2) = 1 - (p(0) + p(1))

    ........... = 1 - (b(0; 20, 0.2465) + b(1; 20, 0.2465))

    ........... ~= 1 - 0.003481 - 20!/[1! 19!] 0.2465^1 (1-0.2465)^19

    ........... ~= 1 - 0.003481 - 0.022773 ~= 0.9737

    RonL
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