# Math Help - Probability question

1. ## 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

2. You missed one term: $P(C\cup D)=P(C)+P(D)-P(C)P(D)$.

3. Haha.. Thank you. It's been a long night