更新1:
答案是 0.0989
機率 PROBABILITY
2007-10-24 9:20 am
an insurance company classifies people as normal or accident prone . Suppose that the probability that a normal person has an accident in a specified year is 0.2 and that for an accident prone person this probability is 0.6 . Further suppose that 18%of the policyholders are accident prone . A policyholder had no accident in a specified year. What is the probability that he or she is accident prone?
回答 (3)
2007-10-24 9:34 am
✔ 最佳答案
18% is accident prone 72% is normal
percentage of holder having no accident in accident prone group
= 18% *0.4
= 7.2 %
percentage of holder having no accident in normal group
= 72% *0.8
= 57.6 %
Therefore, required ans = 7.2 / (7.2+57.6)
= 0.111 = 11.1%
2007-10-25 1:23 am
0.0989
2007-10-24 5:53 pm
P(the person is accident prone)
= P(the person is accident prone and does not have accident) / (P(the person is accident prone and does not have accident) + P(the person is not accident prone and does not have accident))
P(the person is accident prone and does not have accident) = 0.18 x (1-0.6) = 0.072
P(the person is not accident prone and does not have accident) = (1-0.18) x (1-0.2)
= 0.656
Then, P(the person is accident prone) = 0.072 / (0.072 + 0.656) = 0.0989
= P(the person is accident prone and does not have accident) / (P(the person is accident prone and does not have accident) + P(the person is not accident prone and does not have accident))
P(the person is accident prone and does not have accident) = 0.18 x (1-0.6) = 0.072
P(the person is not accident prone and does not have accident) = (1-0.18) x (1-0.2)
= 0.656
Then, P(the person is accident prone) = 0.072 / (0.072 + 0.656) = 0.0989
收錄日期: 2021-04-24 07:48:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071024000051KK00246