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Math Help - Probability question

  1. #1
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    Probability question

    An actuary studying the insurance preferences of automobile owners makes the following conclusions:
    • An automobile owner is twice as likely to purchase collision coverage as disability coverage.
    • The probability that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage.
    • The probability that an automobile owner purchases both collision and disability coverages is 0.15.
    What is the probability that an automobile owner purchases neither collision nor disability coverage?


    Ok, so I have:

    p(c)=2p(d)

    p(c\cap d)=0.15

    =>p(c)*p(d)=0.15

    2p(d)^{2}=0.15

    =>p(d)=\sqrt{0.075}

    p(c)=2p(d)=2\sqrt{0.075}

    p(c'\cap d')=>p(c')*p(d')

    =>(1-2\sqrt{0.075})*(1-\sqrt{0.075})\approx 0.3284

    The above answer is correct but my question is why can't I use De Morgen's law to change p(c'\cap d') to p(c\cup d)' and then just solve  1-p(c \cup d)=>1-(\sqrt{0.075}+2\sqrt{0.075})? Is it because the two events aren't mutually exclusive or something? Thanks
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  2. #2
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    You missed one term: P(C\cup D)=P(C)+P(D)-P(C)P(D).
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  3. #3
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    Haha.. Thank you. It's been a long night
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