Banner Ad 1



No announcement yet.

Help with this question

This topic is closed.
  • Filter
  • Time
  • Show
Clear All
new posts

  • Help with this question

    An insurance company examines its pool of auto insurance customers and gathers the following
    (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

  • #2
    "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


    • #3

      That helps