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# Thread: Collion/Disability Coverage w/ independent probablitiy

1. ## Collion/Disability Coverage w/ independent probablitiy

An actuary studying the insurance preferences of automobile owners makes the following conclusions:
(i) An automobile owner is twice as likely to purchase collision coverage as disability coverage.
(ii) The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage.
(iii) 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?

(A) 0.18
(B) 0.33
(C) 0.48
(D) 0.67
(E) 0.82

I'm not understanding this question too well. I've understood that
P(C)*P(D) = 0.15
P(C) = 2*P(D)
and C has no bearing on D or vice versa

An actuary studying the insurance preferences of automobile owners makes the following conclusions:
(i) An automobile owner is twice as likely to purchase collision coverage as disability coverage.
(ii) The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage.
(iii) 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?

(A) 0.18
(B) 0.33
(C) 0.48
(D) 0.67
(E) 0.82

I'm not understanding this question too well. I've understood that
P(C)*P(D) = 0.15
P(C) = 2*P(D)
and C has no bearing on D or vice versa
I don't know what you don't understand.

Anyway, what you need to find is P[(not C) and (not D)].

First, based on what you have, you can find P(C) and P(D). Then use total probability theorem to find the answer.

Basically, the above tells you how to do the question. If you still do not understand that, go review your textbook or google.

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