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ashi70
August 24th 2005, 11:52 AM
An insurance company examines its pool of auto insurance customers and gathers the following
information:
(i) All customers insure at least one car.
(ii) 70% of the customers insure more than one car.
(iii) 20% of the customers insure a sports car.
(iv) Of those customers who insure more than one car, 15% insure a sports car.
Calculate the probability that a randomly selected customer insures exactly one car and that car
is not a sports car.
A. 0.13 B. 0.21 C. 0.24 D. 0.25 E. 0.30

Ans:- Please help me solve this question. I am having confusion as to when to use Intersection and when to use conditional prob. How should I represent (iv). Is it P(M|C) or P(M[Intersec]C) Where M is the event of insuring multiple cars and C is event of insuring sports car. Also please give me correct answer.
Thanks a bunch

robertr24
August 24th 2005, 02:09 PM
"Of those who insure more than one car, 15% insure a sports car"

== "Given someone insures more than one car, there is a 15% chance they insure a sports car."

As the question states, the P(C|M) = .15. The key word here is "given". Since the P(C int M) = P(C|M) x P(M), the P(C int M) = .15 x .7 = .105.

It can be tricky to know whether to use P(A|B) or P(A int B), depending on the wording. Usually when a question is worded like, "the probability of THIS 'AND' THIS", I'd think it's safe to assume you'd use intersection. But when a question is of the form, "x% of the people in c category are d" or "of the people in c category, x (or x%) are in d", it's conditional, P(D|C). I hope that helps.

As for the question:

M = insures more than one car
C = insures a sports car

P(M' int C') = 1 - P(M union C)

P(M union C) = P(M) + P(C) - P(M int C) = .7 + .2 - .105 = .795

1 - P(M union C) = .205 = .21

ashi70
August 25th 2005, 08:07 AM
That helps