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a probability question
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最佳解答:
P(C didn't get SARS given that B got SARS)=P(C DIDN'T get SARS from SARS carrier B)*P(C DIDN'T got SARS directly from A) (1-4/5)(1-1/8) =7/40 ii)1-P(C got SARS directly from A) =1-4/5 =1/5 2007-04-06 21:49:04 補充： the answer in (i) MUST BE smaller than (ii) C could got SARS directly from A or got from B.IF the question is C didn't get SARS FROM B given that B got SARS, the answer is 7/8
其他解答:
Let P(B) = P(B got SARS directly from A) = 4/5 P(C) = P(C got SARS directly from A) =4/5 P(B|C) = P(B got SARS from C |C got SARS from A) =1/8 P(C|B) = P(C got SARS from B |B got SARS from A) =1/8 (i)find P(C'|B), where C' is the complement of C P(B) = P(C & B) + P(C'&B) P(B) = P(C|B)*P(B) + P(C'|B)*P(B) 1 = P(C|B) + P(C'|B) <--useful equation[divide equation by P(B)] 1 = (1/8) + P(C'|B) P(C'|B) = 7/8 p.s. Since P(B|C)=1/8(not =0), B and C are NOT independent we cannot write P(B)*P(C)=P(B & C) (ii) find P(C'|B') Method 1: first find P(B & C), P(B & C) = P(B|C)*P(C) or P(C|B)*P(B) P(B & C) = (1/8)*(4/5) = 1/10 then find P(B or C), P(B or C) = P(B) + P(C) - P(B & C) P(B or C) = (4/5) + (4/5) - (1/10) = 1.5 <-- impossible (some problem here! Please check if the probabilities given are correct) P(C' & B') = 1 - P(B or C) <-- De Morgan's law P(C' & B') = ... P(C'|B') = P(C' & B') / P(B'), where P(B')=1/5 Method 2: (violate the supposition II) P(B & C) = P(B)*P(C) P(B & C) = (4/5)*(4/5) = 16/25 P(B or C) = P(B) + P(C) - P(B & C) P(B or C) = (4/5) + (4/5) - (16/25) = 24/25 Use De Morgan's law, P(B' & C') = 1 - P(B or C) = 1 - 24/25 = 1/25 P(C'|B') = P(C' & B') / P(B') = (1/25) / (1/5) = 1/5
a probability question
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There were 3 patients A,B and C in the Wellington Hospital.A was a hidden SARS carrier.Suppose:I: P(B got SARS directly from A)=P(C got SARS directly from A)=4/5II: P(B got SARS from C |C got SARS from A) =P(C got SARS from B |B got SARS from A) =1/8(assume there is no way of getting SARS except... 顯示更多 There were 3 patients A,B and C in the Wellington Hospital. A was a hidden SARS carrier. Suppose: I: P(B got SARS directly from A)=P(C got SARS directly from A)=4/5 II: P(B got SARS from C |C got SARS from A) =P(C got SARS from B |B got SARS from A) =1/8 (assume there is no way of getting SARS except between themselves) Find the probability that (i) C didn't get SARS given that B got SARS (ii) C didn't get SARS given that B didn't get SARS最佳解答:
P(C didn't get SARS given that B got SARS)=P(C DIDN'T get SARS from SARS carrier B)*P(C DIDN'T got SARS directly from A) (1-4/5)(1-1/8) =7/40 ii)1-P(C got SARS directly from A) =1-4/5 =1/5 2007-04-06 21:49:04 補充： the answer in (i) MUST BE smaller than (ii) C could got SARS directly from A or got from B.IF the question is C didn't get SARS FROM B given that B got SARS, the answer is 7/8
其他解答:
Let P(B) = P(B got SARS directly from A) = 4/5 P(C) = P(C got SARS directly from A) =4/5 P(B|C) = P(B got SARS from C |C got SARS from A) =1/8 P(C|B) = P(C got SARS from B |B got SARS from A) =1/8 (i)find P(C'|B), where C' is the complement of C P(B) = P(C & B) + P(C'&B) P(B) = P(C|B)*P(B) + P(C'|B)*P(B) 1 = P(C|B) + P(C'|B) <--useful equation[divide equation by P(B)] 1 = (1/8) + P(C'|B) P(C'|B) = 7/8 p.s. Since P(B|C)=1/8(not =0), B and C are NOT independent we cannot write P(B)*P(C)=P(B & C) (ii) find P(C'|B') Method 1: first find P(B & C), P(B & C) = P(B|C)*P(C) or P(C|B)*P(B) P(B & C) = (1/8)*(4/5) = 1/10 then find P(B or C), P(B or C) = P(B) + P(C) - P(B & C) P(B or C) = (4/5) + (4/5) - (1/10) = 1.5 <-- impossible (some problem here! Please check if the probabilities given are correct) P(C' & B') = 1 - P(B or C) <-- De Morgan's law P(C' & B') = ... P(C'|B') = P(C' & B') / P(B'), where P(B')=1/5 Method 2: (violate the supposition II) P(B & C) = P(B)*P(C) P(B & C) = (4/5)*(4/5) = 16/25 P(B or C) = P(B) + P(C) - P(B & C) P(B or C) = (4/5) + (4/5) - (16/25) = 24/25 Use De Morgan's law, P(B' & C') = 1 - P(B or C) = 1 - 24/25 = 1/25 P(C'|B') = P(C' & B') / P(B') = (1/25) / (1/5) = 1/5
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