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Problem 10.14 on Marcel Finan's practice problems

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  • Problem 10.14 on Marcel Finan's practice problems

    Problem 10.14
    Suppose that 25% of all calculus students get an A; and that students who
    had an A in calculus are 50% more likely to get an A in Math 408 as those
    who had a lower grade in calculus. If a student who received an A in Math
    408 is chosen at random, what is the probability that he/she also received
    an A in calculus?
    (Assume all students in Math 408 have taken calculus.)

    --------------------------------------------------------------------------

    The part I do NOT get is the bolded part above. Could someone help me interpret how students who had an A in calculus are 50% MORE LIKELY to get an A in Math 408 as those who had a lower grade in calculus?

    This is the only part that's bugging me. Otherwise, once I understand this part, the rest of the problem should be ok.

    Thanks!

  • #2
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    • #3
      Anyone? ? ?

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      • #4
        Problem 10.14 on Marcel Finan's practice problems

        If:
        Ac : is the event of getting an "A" in calculus
        Acc : is the event of taking other than "A" in calculus, where P(Acc)=1-P(Ac)


        then
        P(Ac) = 0.25
        P(Acc)= 0.75

        Am : is the event of taking an "A" in Math 408
        Acm: is the event of taking a non "A" in Math 408

        P(Am)=1-P(Acm)

        P(Am|Ac) : the probability of getting an "A" in Math 408 if student already has "A" in calculus

        P(Am|Acc) : the probability of getting an "A" in Math 408 for students that have a non "A" in calculus

        NOW : (here is some part that may seem confusing)

        X : is the probability of getting an "A" in Math 408, while having a non"A" in calculus
        (which is P(Am|Acc)=X and X<=1 )

        "the students who had an A in calculus are 50% more likely to get an A in Math 408 as those who had a lower grade in calculus" (ad the base once plus another 50% of the base, where here the base/reference is the probability of getting an A in M 408 while not having A in calculus)
        therefore:

        P(Am|Ac)=P(Am|Acc) + P(Am|Acc) * 50% = X+0.5X = 1.5X

        now:

        P(Am) = P(Am|Ac)P(Ac) + P(Am|Acc)P(Acc)

        P(Am) = 1.5X*0.25+x*0.75 = 0.375X + 0.75X = 1.125X

        We need to find
        "If a student received an "A" in Math 408 is chosen at random, what is the Probability that he/she also received an "A" in calculus?"

        which means: what is the probability of the student of having an A in calculus (also) if he has (for sure) an A in Math 408

        we need to find P(Ac|Am)

        P(Ac|Am) = P(Am|Ac)P(Ac) / P(Am) = 0.375X / 1.125X = 0.375/1.125 = 1/3

        ---
        I hope is correct. That's my take on it.
        vanarpels
        Actuary.com - Newbie Poster
        Last edited by vanarpels; June 22 2012, 08:34 PM.

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