[數學] 有關統計學問題

2009-01-06 12:31 am
In a local cellular phone area, company A accounts for 70% of the cellular phone market, while company B accounts for the remaining 30% of the market. Of the cellular calls made with company A, 2% of the calls will have some sorts of interference, while 3% of the cellular calls with company B will have interference.

Question:
Given that a random selected cellular call is one that has interference, what is the probability it came from company A? (Answer should be between two and four decimal place. eg. 0.12, 0.123, or 0.1234...etc.)

拜託各位統計學高手, 幫我解以上題目好嗎? 謝謝!

回答 (2)

2009-01-07 3:28 am
✔ 最佳答案
你另一條FORECASTING我要多些時間去做呀
WE HAVE
P(A)=0.7,P(B)=0.3
P(Interference|A)=0.02,P(Interference|B)=0.03
So
P(Interference)
=P(A)P(Interference|A)+P(B)P(Interference|B)
=0.7*0.02+0.3+0.03
=0.014+0.009
=0.023
So, by Bayes theorem, the probability it came from company A
=0.014/0.023
=0.6087

2009-01-06 6:54 am
  你可以到這裡問看看

http://www.nice-read.com/Debug_plan/math_debug/math_debug.htm



2009-01-05 23:55:11 補充:
[70%×2%]÷[70%×2%+30%×3%]= 0.609


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