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Math Help - conditional probability problem

  1. #1
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    conditional probability problem

    Hi;
    Not sure where I'm going wrong with this one.

    Given that Susan fails just one exam, find the probability that itsí History she fails.


    maths Pass = 0.7 french pass = 0.8 history pass 0.6



    P(A|B) = P(A ∩ B) / P(B).
    = (0.7 * 0.8 * 0.4) / ((0.7 * 0.8 * 0.4) + (0.7 * 0.2 * 0.4) + (0.3 * 0.2 * 0.4) + (0.3 * 0.8 * 0.4))

    I get 0.56 book gets 0.4956
    cant upload tree diagram sorry.

    Thanks.
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  2. #2
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    Re: conditional probability problem

    conditional probability problem-untitled.jpg
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  3. #3
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    Re: conditional probability problem

    manage to upload tree diagram.
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    Re: conditional probability problem

    Hi Anthonye

    maths Pass = 0.7 french pass = 0.8 history pass 0.6
    Given that Susan fails just one exam, find the probability that itsí History she fails.

    Probability that she fails History but passes both the others is 0.4*0.7*0.8 = 0.224 Just like you had.

    Prob that she fails maths and passes the other 2 = 0.3*0.8*0.6 = 0.144
    Prob that she fails franch and passes the other 2 = 0.2*0.7*0.6 = 0.084

    The probability that she fails one and only one subject = 0.224+0.144+0.084 = 0.452

    Given that Susan fails just one exam, find the probability that itsí History she fails = 0.224 / 0.452 = 0.4956 (4 dec places)
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  5. #5
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    Re: conditional probability problem

    Thanks Melody2 I've sorted it out. these type of questions seem to be written the
    wrong way round to me ie P(B) is what were given but its the first part of the question.
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  6. #6
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    Re: conditional probability problem

    Yes, getting your head around these problems is often the hardest part.
    i am glad that you have it sorted.
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  7. #7
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    Re: conditional probability problem

    Thanks again for all your help.
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  8. #8
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    Re: conditional probability problem

    Another way of looking at it: writing "P" or "F" for pass or fail, and "Math", "French" and "History" in that order, there are 2^3 possible outcomes, with probabilities:
    PPP: (0.7)(0.8)(0.6)= 0.336
    PPF: (0.7)(0.8)(0.4)= 0.224
    PFP: (0.7)(0.2)(0.6)= 0.084
    PFF: (0.7)(0.2)(0.4)= 0.056
    FPP: (0.3)(0.8)(0.6)= 0.144
    FPF: (0.3)(0.8)(0.4)= 0.096
    FFP: (0.3)(0.2)(0.6)= 0.036
    FFF: (0.3)(0.2)(0.4)= 0.024

    Of those, PPF, PFP, and FPP correspond to "fails just one exam" and they have total probability 0.224+ 0.084+ 0.144= 0.452.
    That one exam being French is PFP which has probability 0.084 so the probability that she will fail French given that she fails just one exam is \dfrac{0.084}{0.452}= 0.1858
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  9. #9
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    Re: conditional probability problem

    Quote Originally Posted by HallsofIvy View Post
    Another way of looking at it: writing "P" or "F" for pass or fail, and "Math", "French" and "History" in that order, there are 2^3 possible outcomes, with probabilities:
    PPP: (0.7)(0.8)(0.6)= 0.336
    PPF: (0.7)(0.8)(0.4)= 0.224
    PFP: (0.7)(0.2)(0.6)= 0.084
    PFF: (0.7)(0.2)(0.4)= 0.056
    FPP: (0.3)(0.8)(0.6)= 0.144
    FPF: (0.3)(0.8)(0.4)= 0.096
    FFP: (0.3)(0.2)(0.6)= 0.036
    FFF: (0.3)(0.2)(0.4)= 0.024

    Of those, PPF, PFP, and FPP correspond to "fails just one exam" and they have total probability 0.224+ 0.084+ 0.144= 0.452.
    That one exam being French is PFP which has probability 0.084 so the probability that she will fail French given that she fails just one exam is \dfrac{0.084}{0.452}= 0.1858
    Ooops! I misread the problem! It was "History", not "French". That will be PPF which has probability \dfrac{0.224}{0.452}= 0.4956, the same as Melody2 said. Whew!
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  10. #10
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    Re: conditional probability problem

    Hi HallsofIvy,
    I was a bit confused when your name came up.
    Nice to meet you again!
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