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.

Rating

+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)).

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