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Homework Help: Class and student probability question

  1. Jan 24, 2010 #1
    student A comes to 80% of the classes while student B comes to 60%, (A,B are independant of one another), on a given day exactly one student arrived, what is the probability that it was student A


    P(A)=0.8
    P(B)=0.6

    P([tex]\bar{A}[/tex])*P(B)=0.12
    P([tex]\bar{B}[/tex])*P(A)=0.32

    C=> only one student arrived
    P(C)=P([tex]\bar{A}[/tex])*P(B) +P([tex]\bar{B}[/tex])*P(A)=0.12 + 0.32 =0.44
    P(C/A)=P([tex]\bar{B}[/tex])*P(A)=0.32

    P(A/C)=[tex]\frac{P(A)*P(C/A)}{P(C)}[/tex]=[tex]\frac{0.8*0.32}{0.44}[/tex]=0.58

    this is wrong the correct answer is meant to be 0.73, any help??
     
  2. jcsd
  3. Jan 24, 2010 #2

    Dick

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    Re: probability

    Why isn't it just P(A)*P(not B)/(P(A)*P(not B)+P(B)*P(not A))? Why the extra P(A) factor?
     
  4. Jan 24, 2010 #3
    Re: probability

    apparently it is, but i know P(A/C) is the probability of A if i know that C has occured, and i thought that that was what i needed to use here
     
  5. Jan 24, 2010 #4

    Dick

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    Re: probability

    The numerator is P(A|C). So the whole answer is P(A|C)/P(C). I'm questioning where the extra P(A) came from?
     
  6. Jan 24, 2010 #5
    Re: probability

    as far as i can tell im looking for P(A|C)-> the probability of A if C has occurred,

    P(A|C)= P(A)*P(C|A)/P(C)

    http://en.wikipedia.org/wiki/Bayes'_theorem
     
  7. Jan 24, 2010 #6

    Dick

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    Re: probability

    Ok, if you want to do it that way, then P(C|A) is the probability that A is there and B is not there divided by probability that A is there. That's 0.32/0.8. You are drawing a Venn type diagram of this, right?
     
  8. Jan 24, 2010 #7

    vela

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    Re: probability

    If you know A arrived, the probability that only one student arrived is the probability that B didn't, so P(C|A)=P(not B).
     
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