xzero_1989216 asked in 科學數學 · 1 decade ago

有誰可以教我這題? (條件機率/貝式網路)

At a certain stage of a criminal investigation, the inspector in charge is 60% convinced of the guilt of a certain suspect. Suppose now that a new piece of evidence that shows that the criminal has a certain characteristic (such as left-handedness, baldness, brown hair, etc.) is uncovered. If 20% of the population possesses this characteristic, how certain of the guilt of the suspect should the inspect now be ifit turns out that the suspect is among this group?

在調查間段時,檢察官已有60%的把握機率確定一位嫌疑人是犯人‧然後一份新的報告指出犯人的外冒特徵‧假設20%的人口擁有這些特徵而那位嫌疑人也有這些特徵,那請問檢察官現在有幾成的把握確定那位嫌疑人就是犯人?

Let us now suppose that the new evidence is subject to different possible interpretations, and in fact only shows that it is 90% likely that the criminal possesses this certain characteristic. In this case, how likely would it be that the suspect is guilty (assuming, as before, that he has this characteristic)?

假如之前的報告所指出的特徵只有90%是犯人擁有的特徵,那請問檢察官有幾成的把握確定那位嫌疑人就是犯人?

Update:

第一題

A指他有沒有罪 B指他有沒有犯人的特徵

他有罪的機率是60% 無罪的機率是 40%

如果他有罪那他有100%的機率有犯人的特徵,而如果他沒有罪那他有20%機率有犯人的特徵‧

題目給了 他有犯人的特徵

那他有罪的機率是 P(A and B) / P(B) = (0.6 x 1) / (0.6 x 1 + 0.4 x 0.2) = 0.88235

我的想法是對的嗎?

1 Answer

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  • 1 decade ago
    Favorite Answer

    第一題

    以A表示該嫌犯有罪的事件,以B表示該嫌犯有犯人特徵的事件。我們要計算 「給定該嫌犯有犯人特徵,他有罪的機率是多少?」,也就是 P(A | B)。

    P(A) = 0.6, P(AC) = 0.4,

    P(B | A) = 1,也就是「給定他有罪,那他有100%的機率有犯人的特徵」。

    P(B | AC) = 0.2,也就是「給定他無罪,那他有20%的機率有犯人的特徵」。

    圖片參考:http://img364.imageshack.us/img364/8053/0805241pu8...

    第二題

    P(A) = 0.6, P(AC) = 0.4,P(B | A) = 0.9,P(B | AC) = 0.2

    圖片參考:http://img249.imageshack.us/img249/1095/0805242uy7...

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