Probability

+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)).

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

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