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Thread: Probability Question (Ambiguous)

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    Red face Probability Question (Ambiguous)




    SOLVED BY MYSELF, YOU CAN CLOSE THE THREAD

    Hi guys,

    I'm new to this forum, and some of you might find it annoying that I post a question straight off, but the assignment is due tomorrow and I have two different answers. I do promise to keep posting in threads (when I have something to say)

    It's basic probability. Here goes:

    Insurance company X sells a number of different policies, among which 60% are for drivers, 40% for homeowners and 20% for both types of policies. Let A1 be people with only a motor policy, A2 those with only a homeowner policy, A3 people with both types and A4 those with other types of policies.

    (i) Assume a person is selected at random. What is the probability that this person belongs to A1/A2/A3/A4 ? (basically they want us to get probability of each separately).

    (ii) Let B denote the event that a policyholder will renew at least one of the car or home insurance policies. Based on past experience, we can assume that

    P(B|A1) = 0.6,
    P(B|A2)= 0.7,
    P(B|A3)= 0.8

    Given that the person selected at random has a car or a home insurance policy, what is the probability that this person will renew at least one of those policies?

    (It was I who made some bits italic and bold).


    My Solution

    (i) Basic Venn diagram, get
    P(A1)=0.4
    P(A2)=0.2
    P(A3)=0.2
    P(A4)=0.2

    (ii) First of all, I assume that you're given that the person is allowed to have both policies (A3), because it says "at least one".

    From the given information, I worked out
    P(B ∩ A1)=0.24
    P(B ∩ A2)=0.14
    P(B ∩ A3)=0.16

    Solution 1

    I drew a tree diagram (numbers in brackets represent probabilities)
    • A1 (0.4) ----> B (0.6) so total prob.= 0.24
    • A2 (0.2) ----> B(0.7) so total prob. = 0.14
    • A3 (0.2) ----> B(0.8) so total prob. = 0.16

    Then I just added the "total" probabilities to get 0.54

    Solution 2

    I said that I'm looking for P(B | (A1 U A2 U A3))

    which equals to P(B ∩ (A1 U A2 U A3)) / P(A1 U A2 U A3)

    which equals to (0.24 + 0.16 + 0.14) / (0.4 + 0.2 + 0.2)

    which is 0.675

    Which one is right? And what's the difference between the two approaches?

    Thanks!
    Last edited by cluborange; 10-24-2013 at 05:03 PM. Reason: Solved

  2. #2
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    Re: Probability Question (Ambiguous)


    Problem found, first solution is wrong, we're GIVEN that it's either A1 or A2 or A3, so multiplying P(b) by each of P(A)s doesn't make sense

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