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[問題]費費 06-01

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[問題]費費 06-01

帖子cheyen » 2005-12-06 00:13

In an insurance company, each policy has a paper record or an electric record, or both of them.
60 percent of the policies having incorrect paper record have incorrect electric record and
75 percent of the policies having incorrect electric record have incorrect paper record.
3 percent of all the policies have both incorrect paper and incorrect electric records.

If we randomly pick out one policy, what is the probability that it is one having both correct paper and correct electric record?


集合的相關問題
請問這題有人做過嗎?
答案是94%
小弟腦袋瓜沒通,麻煩大將幫我解圍下 :laugh
多謝大家 感恩感恩
cheyen
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Re: [問題]費費 06-01

帖子zazer.lin » 2005-12-06 12:49

設 paper record 正確的機率為P(p), 則不正確的機率為 P(p[sup]c[/sup]) = 1 - P(p)
設 electric record 正確的機率為P(e), 則不正確的機率為 P(e[sup]c[/sup]) = 1 - P(e)

依據題意:
60 percent of the policies having incorrect paper record have incorrect electric record

P(e[sup]c[/sup]|p[sup]c[/sup]) = 60% = 0.6
==> P(e[sup]c[/sup] 交集 p[sup]c[/sup]) / P(p[sup]c[/sup]) = 0.6
==> 0.6 * P(p[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]).........<I>

75 percent of the policies having incorrect electric record have incorrect paper record

P(p[sup]c[/sup]|e[sup]c[/sup]) = 75% = 0.75
==> P(p[sup]c[/sup] 交集 e[sup]c[/sup]) / P(e[sup]c[/sup]) = 0.75
==> 0.75 * P(e[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]).........<II>

3 percent of all the policies have both incorrect paper and incorrect electric records

P({p 交集 e}[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 3% = 0.03.........<III>
並且 P({p 交集 e}[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 1 - P({p 聯集 e}) = 0.03
==> P({p 聯集 e}) = 0.97

由<I><III>可得知
0.6 * P(p[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 0.03
==> 0.6[1 - P(p)] = 0.03
==> P(p) = 0.95

由<II><III>可得知
0.75 * P(e[sup]c[/sup]) = P(p[sup]c[/sup] 交集 e[sup]c[/sup]) = 0.03
==> 0.6[1 - P(e)] = 0.03
==> P(e) = 0.96

題目問的是
If we randomly pick out one policy, what is the probability that it is one having both correct paper and correct electric record

意即求 P(p 交集 e) = ?
由貝式定理得知
P({p 聯集 e}) = P(p) + P(e) - P(p 交集 e)
0.97 = 0.95 + 0.96 - P(p 交集 e)
P(p 交集 e) = 0.94

這題題目有點繞...把文式圖畫出來應該會比較好理解... ;))
出國留學真是條不歸路!!
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zazer.lin
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[問題]費費 06-01

帖子cheyen » 2005-12-06 19:59

多謝zazer.lin相救
你寫的很仔細
你的解法真讓我想起我以前高中老師style
去咀嚼一下
不懂再上來提問
感恩阿 :laugh
cheyen
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中級會員
 
帖子: 135
注册: 2004-11-02 11:54


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