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Anonymous
Anonymous asked in 科學及數學數學 · 10 years ago

Probability

5% of the people in a city is Hepatitis B carrier. Probability that a diagnostic test gives a correct diagnosis is 0.98. If a man is diagnosed to be a carrier, what probability that the man is actually not a carrier?

Update:

To : 福爾摩斯. Most people think in the same way as you but in fact you are wrong. Adam is correct, please study his answer.

3 Answers

Rating
  • Adam
    Lv 7
    10 years ago
    Favorite Answer

    +ve: diagnosed to be a carrier

    -ve: diagnosed to be not a carrier

    P(CARRIER) = 5%

    P(not CARRIER) = 1 - 5%

    P(+ve|CARRIER) = P(-ve|notCARRIER) = 0.98

    P(-ve|notCARRIER) = P(+ve|CARRIER) = 1 - 0.98

    P(CARRIER and +ve)

    = P(CARRIER) x P(+ve|CARRIER)

    = 5% x 0.98

    = 0.049

    P(notCARRIER and +ve)

    = P(notCARRIER) x P(+ve|notCARRIER)

    = (1 - 5%)(1 - 0.98)

    = 0.019

    P(+ve)

    = P(CARRIER and +ve) + P(notCARRIER and +ve)

    = 0.049 + 0.019

    = 0.068

    P(notCARRIER|+ve)

    = P(notCARRIER and +ve) / P(+ve)

    = 0.019/0.068

    = 19/68

    (or 0.2794 (4 sig.fig.))

    2010-10-04 11:34:27 補充:

    To: 福爾摩斯

    If a person is randomly choosen, P(not a carrier) = 1 - 0.98 = 0.02

    However, the person is diagnosed to be a carrier, but not randomly choosen.

    Now, we are calculating P((not a carrier) | (diagnosed to be a carrier)).

    Source(s): adam, adam
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  • 10 years ago

    To me, the first sentence is a trap.

    Since the probability of a diagnostic test giving a correct diagnosis is 0.98,

    if a man is diagnosed to be a carrier, the probability that the man is actually not a carrier is simply 1- 0.98 = 0.02

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  • 10 years ago

    5% of the people in a city is Hepatitis Bcarrier. Probability that a diagnostic test gives a correct diagnosis is 0.98.If a man is diagnosed to be a carrier, what probability that the man isactually not a carrier?SolA:為Hepatitis帶原者事件B:不為Hepatitis帶原者事件C:測試為Hepatitis帶原者事件P(A)=0.05,P(B)=1-0.05=0.95P(C|A)=0.98P(B|C)=P(BC)/P(C)=P(BC)/[P(BC)+P(AC)]=P(C|B)*P(B)/[P(C|B)*P(B)+P(C|A)*P(A)]=P(C|B)*0.95/[P(C|B)*0.95+0.98*0.05]條件不足

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