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In a class of 75 students,15 are above average,45 are average and the rest below average achievers.The probability that an above average achieving student fails is 0.005,then an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15.If a student is known to have passed,what is the probability that he is a below average achiever?

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Susan

  1. let O be over average, A be average, and B be below average, F for fail and P for pass
    prob(O) = 15/75 = 1/5
    prob(A) = 45/75 = 3/5
    prob(B) = 15/45 = 1/5
    prob(P) = x
    prob(F) = 1-x , not really needed

    so we have
    AF = .05
    AP = (3/5)x
    OF = .005
    OP = (1/5)x
    BF = .15
    BP = (1/5)x

    .05 + 3x/5 + .005 + x/5 + .15 + x/5= 1
    x = 1 - .05 - .005 - .15 = .795

    Thus:
    AF = .05
    AP = .477
    OF = .005
    OP = .159
    BF = .15
    BP = .159 , notice they add up to 1

    So we have BP = .159
    B(.795) = .159
    B = .159/.795 = .2

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    Reiny

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