✔ 最佳答案
Ei, i = 1, 2, 3 denote the event that judge i casts a guilty vote.
Suppose that when the defendant is, in fact, guilty, each judge will independently vote guilty with probability 0.7, ...
由以上敘述可知: E1, E2, E3 是 conditionally independent,
given the defendant is guilty or innocent.
令 A denote the event that the defendant is guilty.
而 A' 是 A 的餘事件, 即 defendant is innocent.
則:
P(Ei|A) = 0.7, P(Ei|A') = 0.2, i=1,2,3.
又: P(A) = 0.7.
(a) P(E3|E1∩E2) = P(E1∩E2∩E3)/P(E1∩E2)
= (0.7*0.7^3+0.3*0.2^3)/(0.7*0.7^2+0.3*0.2^2)
= .2425/.355 = 97/142
其中, 以 P(E1∩E2) 為例, 其計算方法是
P(E1∩E2)=P(A)P(E1∩E2|A)+P(A')P(E1∩E2|A')
= P(A)P(E1|A)P(E2|A) + P(A')P(E1|A')P(E2|A')
= 0.7*0.7^2 + 0.3 * 0.2^2
(b) P(E3|(E1∩E2')∪(E1'∩E2))
= P(E3∩[(E1∩E2')∪(E1'∩E2)])/P((E1∩E2')∪(E1'∩E2))
P(E3∩[(E1∩E2')∪(E1'∩E2)])
= 0.7*0.7*(2*0.7*0.3) + 0.3*0.2*(2*0.2*0.8)
= .2058 + .0192 = .2250
P((E1∩E2')∪(E1'∩E2))
= 0.7*(2*0.7*0.3) + 0.3*(2*0.2*0.8)
= .390
故 P(E3|(E1∩E2')∪(E1'∩E2)) = .2250/.390 = 15/26
(c) P(E3|E1'∩E2') = P(E3∩E1'∩E2')/P(E1'∩E2'),
其中
P(E3∩E1'∩E2')= 0.7*0.7*0.3^2+0.3*0.2*0.8^2
= .0441 + .0384 = .0825
P(E1'∩E2') = 0.7*0.3^2+0.3*0.8^2 = .063+.192 = .255
故 P(E3|E1'∩E2') = .0825/.255 = 33/102.
注意 P(Ei) = 0.7*0.7+0.3*0.2 = 0.55 , i=1,2,3
而 P(E1∩E2) = .355 ≠ P(E1)P(E2), 故 E1 與 E2 不獨立,
對 E2與E3, E1與E3 結果亦同. 因此, E1, E2, E3 兩兩不獨立,
踵然 E1, E2, E3 conditionally independent(given A or A').
2011-03-29 21:08:51 補充：
P(E1∩E2)=P(A)P(E1∩E2|A)+P(A')P(E1∩E2|A')
= P(A)P(E1|A)P(E2|A) + P(A')P(E1|A')P(E2|A')
= 0.7*0.7^2 + 0.3 * 0.2^2
P(juge 1 & 2 都判有罪)
= P(確實有罪, 且被 juge 1 & 2 判有罪) + P(無罪, 卻被 juge 1 &2 判有罪)
2011-03-29 21:09:17 補充：
P(juge 1 & 2 都判有罪)
= P(確實有罪)P(juge 1 & 2 都判有罪|defendant 有罪)
+ P(其實無罪)P(juge 1 & 2 都判有罪| defwndant 無罪)
= P(確實有罪)*0.7^2 + P(其實無罪)*0.2^2
= 0.7*0.7^2 + 0.3*0.2^2
2011-03-29 21:13:23 補充：
令 E 是任一事件 (如 E1, E2, E3, E1∩E2, E1∩E2∩E3),
A 也是一事件. 則可以用 A 把事件 E 做分割, 即
P(E) = P(E∩A) + P(E∩A')
再把 P(E∩A) = P(A)P(E|A) 及 P(E∩A') = P(A')P(E|A') 代入, 得
P(E) = P(A)P(E|A) + P(A')P(E|A').
(上式也稱 "全機率定理". 你既知 Bayes 定理, 也應知這公式.)
