3.指出下列等式和命题是否成立,并说明理由:-|||-(1) cup B=(AB)cup B;-|||-(2) overline (A)B=Acup B:-||
1.证明:-|||-(1) (Acup B)|C=A|(Bcup C) ;-|||-(2) (Acup B)|C=(A|C)cup (B|C) -
5.化简:-|||-(1)(AB∪C)(AC);-|||-(2) (Acup B)(Acup overline (B)).
A,B为两事件,若(Acup B)=0.8,(Acup B)=0.8,则()成立.A.(Acup B)=0.8B.(Acup B)=0.8C.(Acup B)=
设A、B、C是任意三个随机事件,则下列命题正确的是( )(Acup B)-B=A-B(Acup B)-B=A-B(Acup B)-B=A-B(Acup B
A,B,C是任意事件,在下列各式中,不成立的是()(A) (A-B)cup B=Acup B.-|||-(B) (Acup B)-A=B.-|||-(C) (A
(1)设事件A与B相容,则有 ()-|||-(A) (Acup B)=P(A)+P(B);-|||-(B) (Acup B)=P(A)+P(B)-P(AB);-
[题目]证明 cup (Bcap C)=(Acup B)cap (Acup C)
(1)(overline(AB)cup C)(overline(AC)); (2)(Acup B)(Acupoverline(B)).6.证明:(Acup B
二、化简下列各式-|||-1、 (Acup B)(overline (A)cup B)(Acup overline (B))