5.单远题-|||-下列等式不成立的是 ()-|||-(A) cup B=(Aoverline (B))cup B-|||-(B) -B=Aoverline (B) .-|||-(C) =(AB)cup (Aoverline (B)) .-|||-(D) (A-B)cup B=A

1.下列等式不成立的是 () .-|||-(A) =ABcup Aoverline (B); (B) -B=Aoverline (B);-|||-(C) (ov

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

设 A，B，C 为三个事件，指出下列各等式成立的条件.(1) ABC = A(2) A cup B cup C = A(3) A cup B = AB(4) A

已知 (overline (A))=0.3 =(Aoverline (B))=0.4 ,则 (overline (A)cup overline (B))=-||

10.证明下列事件的运算公式:()-|||-(1) =ABcup Aoverline (B); __-|||-(2) cup B=Acup overline (

10.证明下列事件的运算公式:-|||-(1) =ABcup Aoverline (B);-|||-__-|||-(2) cup B=Acup overline

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:-||

5.设A,B,C为三个随机事件,且 (overline (A)cup overline (B))=0.8 (overline (A)cup overline (

用作图法说明下列各命题成立:(1)cup B=(A-AB)cup B，且右边两事件互斥;(2)cup B=(A-AB)cup B,且右边三事件两两互斥。用作图法

2.1.2 用逻辑代数定律证明下列等式:-|||-(1) +overline (A)B=A+B-|||-__-|||-(2) +Aoverline (B)C+A