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Thread: Marcel Q11.15

  1. #1
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    Marcel Q11.15

    Hello guys:
    Problem 11.15
    Among undergraduate students living on a college campus, 20% have an
    automobile. Among undergraduate students living off campus, 60% have an
    automobile. Among undergraduate students, 30% live on campus. Give the
    probabilities of the following events when a student is selected at random:
    (a) Student lives off campus
    (b) Student lives on campus and has an automobile
    (c) Student lives on campus and does not have an automobile
    (d) Student lives on campus and/or has an automobile
    (e) Student lives on campus given that he/she does not have an automobile.

    My attempt:

    I started by drawing the table below. Where C=campus C`=off-campus A=automobile A`= No automobile.

    C C`

    A 0.20 0.60

    A` 0.10 0.10



    I don't see any conditional statement in the question except for (e).
    Last edited by Dabsy; May 8th 2013 at 05:08 PM.

  2. #2
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    Are you asking on how to do these or just commenting?

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    I'm asking, why are we using conditional statement in the solution? or please help me with the solution then I can ask my questions. Thanks.

  4. #4
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    Quote Originally Posted by Dabsy View Post
    I'm asking, why are we using conditional statement in the solution? or please help me with the solution then I can ask my questions. Thanks.
    I don't see a solution so I can't answer why "we" are using conditional statements. Show your solution and if you are not getting the correct answer, we can show you where your logic is potentially flawed.

  5. #5
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    Ok, below is a solution not mine,but in the question to me there is nothing suggesting conditional statement,that's why I'm confused. I started solving the problem;which I gave in the table above, but got wrong answers,except the first one.
    We are given: P(A|C) = 0.2 P(A|C') = 0.6 P(C) = 0.3 and P(C') = 0.7

    So, P(A)=P(A|C) P(C) + P(A|C') P(C') = 0.48

    Then, P(AC) = P(A|C) P(C) = 0.06

    Now I am confused by the wording.

    The probability that a student lives on campus AND has an automobile would be P(AC) = 0.06.

    The probability that a student lives on campus OR has an automobile would be P(AUC) = P(A) + P(C) - P(AC) = 0.72

    So I guess the question is referring to P(AUC) = 0.72

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