P(B|A3)=02,求P(A1|B)A1间互不相容,B能且只能与A1中之一同时发生,由 贝叶斯公式得 P(B AP(AD) P(A1\BP(B|4)P(A)+P(B|42)PA2)+P(B|4)P(43) 0.05×0 0.0435 0.05×0.1+0.1×0.7+0.2×0.22 15、解:记事件“发AAAA”为A4,事件“发BBBB”为B,事件“发CCCC”为C, 事件“收ABCA”为D,则P(A4)=0.3,P(B4)=0.4,P(C4)=0.3,为求P(D|A4) 考虑到发AAAA,而收到ABCD,有两个字母被准确收到,另两个字母被误收,故 P(D|4)=0.62×0.22=00144。同理可求得P(D|B)=P(D|A2)=06×0.2 00048,欲求的概率是P(A4|D),而事件A4,B4,C间两两互不相容,又D能且 只能与A4,B,C4之一同时发生,由贝叶斯公式得欲求的概率为 P(AD) P(A)P(D A" P(A4)P(D|A4)+P(B+)P(D|B4)+P(C4)P(D|C4) 0.3×0.0144 0.5625 .3×0.0144+04×0.0048+0.3×0.004816 16、证 (1)P((AUB)nC)=P(ACU BC)=P(AC)+P(BC)-P(ABC) P(AP(C)+P(B)P(C)-P(A)P(B)P(C) P(C)[P(A)+ P(B)-P(AB)]=P(C)P(AUB AUB与C独立。 2)P(ABC)= P(A)P(B)P(C)=P(AB)P(C) AB与C独立。 (3) P((A-B))=P(ABC)=P(AC(Q2-B))=P(AC)-P(ABC) P(A)P(C)-P(A)P(B)P(C) P(CIP(A)-P(AB)= P(C)P(A-B), A-B与C独立 17, E: P(AB)=P(AUB)=1-P(AUB)=1-[P(A)+P(B)-PAB)] 1-P(A)-P(B)+P(A)P(B)=(1-P(A)(1-P(B) =P(A)P(B), 理可证P(AC)=P(A)P(C) P(BC)=P(B)P(C) 又有 P(ABC)=P(AUBUC)=1-P(AUBUC) 1-P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC)P(B | A3 ) = 0.2 ,求 ( | ) P A1 B 。 Ai 间互不相容,B 能且只能与 Ai 中之一同时发生,由 贝叶斯公式得 ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | ) 1 1 2 2 3 3 1 1 1 P B A P A P B A P A P B A P A P B A P A P A B + + = 0.0435 23 1 0.05 0.1 0.1 0.7 0.2 0.2 0.05 0.1 = = + + = . 15、解:记事件“发 AAAA”为 A4,事件“发 BBBB”为 B 4,事件“发 CCCC”为 C 4, 事件“收 ABCA”为 D,则 ( ) 0.3, ( ) 0.4, ( ) 0.3, 4 4 4 P A = P B = P C = 为求 ( | ) 4 P D A , 考虑到发 AAAA,而收到 ABCD,有两个字母被准确收到,另两个字母被误收,故 ( | ) 0.6 0.2 0.0144 4 2 2 P D A = = 。同理可求得 4 4 3 P(D | B ) = P(D | A ) = 0.60.2 = 0.0048 ,欲求的概率是 ( | ) 4 P A D ,而事件 4 4 4 A ,B ,C 间两两互不相容,又 D 能且 只能与 4 4 4 A ,B ,C 之一同时发生,由贝叶斯公式得欲求的概率为 ( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | ) ( | ) 4 4 4 4 4 4 4 4 4 P A P D A P B P D B P C P D C P A P D A P A D + + = 0.5625 16 9 0.3 0.0144 0.4 0.0048 0.3 0.0048 0.3 0.0144 = = + + = . 16、证: (1) P((A B) C) = P(AC BC) = P(AC) + P(BC) − P(ABC) = P(A)P(C) + P(B)P(C) − P(A)P(B)P(C) = P(C)[P(A) + P(B) − P(AB)] = P(C)P(A B) , ∴ A B 与 C 独立。 (2) P(ABC) = P(A)P(B)P(C) = P(AB)P(C) ∴AB 与 C 独立。 (3) P((A− B)C) = P(ABC) = P(AC( − B)) = P(AC) − P(ABC) = P(A)P(C) − P(A)P(B)P(C) = P(C)[P(A) − P(AB)] = P(C)P(A − B) , ∴ A − B 与 C 独立。 17、证: P(AB) = P(A B) = 1− P(A B) = 1−[P(A) + P(B) − PAB)] = 1− P(A) − P(B) + P(A)P(B) = (1− P(A))(1− P(B)) = P(A)P(B) , 同理可证 P(AC) = P(A)P(C), P(BC) = P(B)P(C) . 又有 P(ABC) = P(A B C) = 1− P(A B C) =1−P(A)+ P(B)+ P(C)− P(AB)− P(AC)− P(BC)+ P(ABC)