overline (C)+AB-|||-D . C+AB

=AB+overline (AC)+overline (BC)+overline (C)D+overline (D) __-|||-__ __

化简：overrightarrow((overline{AB) cup C) (overline(AC))} 化简：$\overrightarrow{(\ove

一.化简overline (AB)cup C)(overline (AC))一.化简

(1)(overline(AB)cup C)(overline(AC)); (2)(Acup B)(Acupoverline(B)).6.证明:(Acup B

(D)C与AB独立.A、AB、BC、CD、DA、AB、BC、CD、D

(A)P(overline(AB))=0; (B)P(AB)=P(A.)P(B.); (C.)P(A)=1-P(B); (D.)P(overline(A)c

4.设A,B,C为随机事件,且A,C互不相容,P(AB)=(1)/(2),P(C)=(1)/(3),求P(AB|overline(C)).4.设A,B,C为随机

2.设A,B为二随机事件,则 (ABcup Aoverline (B))(Acup overline (A)overline (B))= ()-|||-(A)A

2.设A,B为二随机事件,则 (ABcup Aoverline (B))(Acup overline (A)overline (B))= ()-|||-(A)A

1.已知事件A,B,C,则以下结论错误的是 () .-|||-(a) overline (AB)=overline (A)cup overline (B) (b