已知函数 $F = ABCD + \overline{ABC} + \overline{ABCD} + \overline{BCD}$，约束条件为 $\overline{B} \cdot \overline{C} + \overline{BD} = 0$，将其化简为最简“与-或”式为()。
A $\overline{B} + \overline{A} \cdot \overline{C} + \overline{A} \cdot \overline{D} + \overline{C} \cdot \overline{D}$
B $\overline{B} + \overline{A} \cdot \overline{C} + \overline{A} \cdot \overline{D} + \overline{C} \cdot \overline{D}$
C $\overline{B} + \overline{A} \cdot \overline{C} + \overline{A} \cdot \overline{D} + \overline{C} \cdot \overline{D}$
D $\overline{B} + \overline{A} \cdot \overline{C} + \overline{A} \cdot \overline{D} + \overline{C} \cdot \overline{D}$
函数=overline (A)(B+overline (C)cdot overline (DE)) 的反函数是-|||-__ __（ ）A、=overline
[题1.12]用图形法将下列函数化简成为最简与或式:-|||-(1) _(1)=overline (A)BCD+overline (A)BCD+overline
5.设A,B,C为三个随机事件,且 (overline (A)cup overline (B))=0.8 (overline (A)cup overline (
设A，B，C为三个事件，且P(overline(A)cup overline(B))=0.9，P(overline(A)cup overline(B)cup o
[题目]-|||-已知 (overline (A))=0.3, (overline (B))=0.4P(overline (AB))=0.5, 求条件概率-||
=AB+overline (AC)+overline (BC)+overline (C)D+overline (D) __-|||-__ __
设A，B，C为三个事件，且(overline (A)cup overline (B))=0.9-|||-__ __，(overline (A)cup overl
已知 (overline (A))=0.3 =(Aoverline (B))=0.4 ,则 (overline (A)cup overline (B))=-||
已知P（A）=0.4，P（B）=0.3，P（B|overline(A)）=0.4．求：（1）P（overline(A)B）；（2）P（overline(A)ov
设A，B，C是三个随机事件，则Aoverline(B)overline(C)∪overline(A)Boverline(C)∪overline(A)overli