1.证明:-|||-(1) (Acup B)|C=A|(Bcup C) ;-|||-(2) (Acup B)|C=(A|C)cup (B|C) -
(1)(overline(AB)cup C)(overline(AC)); (2)(Acup B)(Acupoverline(B)).6.证明:(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
二、化简下列各式-|||-1、 (Acup B)(overline (A)cup 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)=
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
5.化简:-|||-(1)(AB∪C)(AC);-|||-(2) (Acup B)(Acup overline (B)).
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:-||
[题目]-|||-设随机事件A、B、C两两互斥,且 (A)=0.2,-|||-(B)=0.3, (C)=0.4, 则 (Acup B-C)= __
10.证明下列事件的运算公式:()-|||-(1) =ABcup Aoverline (B); __-|||-(2) cup B=Acup overline (