用画图的方式求解概率论题目很方便，但难点在于如何画和怎么理解

二、解析

解法一：画图

\begin{aligned} & AB = \bar{A} \bar{B} \\ \Rightarrow & \textcolor{springgreen}{\boldsymbol{ A \cup B = \Omega }} \end{aligned}

解法二：特例猜测

$$\textcolor{yellow}{ \begin{cases} A = \bar{A} \\ B = \bar{B} \end{cases} }$$

$$\textcolor{yellow}{ \begin{cases} A = \bar{B} \\ B = \bar{A} \end{cases} }$$

$$\textcolor{orangered}{ \begin{cases} A \neq \bar{A} \\ B \neq \bar{B} \end{cases} }$$

$$\textcolor{springgreen}{ \begin{cases} A = \bar{B} \\ B = \bar{A} \end{cases} }$$

\begin{aligned} & \textcolor{springgreen}{ A \cup B } \\ = & A \cup \bar{A} \\ = & \textcolor{springgreen}{\boldsymbol{ \Omega }} \end{aligned}

解法三：借助摩根律

\begin{aligned} A B \\ & = \bar{A} \cap \bar{B} \\ & = \overline{A \cup B} \end{aligned}

$$\textcolor{yellow}{ AB \subset (A \cup B) }$$

\begin{aligned} A \cup B \\ & = ( A \cup B ) \cup (\textcolor{orange}{A B}) \\ & = ( A \cup B ) \cup (\textcolor{orange}{\bar{A} \cap \bar{B}}) \\ & = ( A \cup B ) \cup (\textcolor{orange}{\overline{A \cup B}}) \\ & = \textcolor{springgreen}{\boldsymbol{\Omega}} \end{aligned}

解法四：用事件概率的思想计算

$$\textcolor{yellow}{ P(B – A) = P(B) – P(AB) } \tag{1}$$

$$AB \subset A$$

$$\textcolor{orange}{ P(B – A) = P(B) – P(A) } \tag{2}$$

\begin{aligned} & \textcolor{springgreen}{P(\bar{A} B)} \\ = & P[(1-A) B] \\ = & P (B – AB) \\ \xlongequal{AB \subset A} & \textcolor{springgreen}{P(B) – P(AB)} \end{aligned} \tag{3}

\begin{aligned} & \textcolor{springgreen}{P(A \bar{B})} \\ = & P[A (1-B)] \\ = & P (A – AB) \\ \xlongequal{AB \subset A} & \textcolor{springgreen}{P(A) – P(AB)} \end{aligned} \tag{4}

$$\textcolor{yellow}{ P(AB) = P(\bar{A} \bar{B}) }$$

\begin{aligned} & \begin{cases} \textcolor{magenta}{P(AB) = P(A) – P(A \bar{B})} \\ P(\bar{A} \bar{B}) = \textcolor{magenta}{P(\bar{A}) – P(\bar{A} B)} \end{cases} \\ \Rightarrow & P(AB) = P (\bar{A} \bar{B}) \\ \Rightarrow & \textcolor{magenta}{2} P(AB) = \textcolor{orange}{P(A) + P(\bar{A})} – P(A \bar{B}) – P(\bar{A} B) \\ \Rightarrow & 2P(AB) = \textcolor{orange}{1} – P(A \bar{B}) – P(\bar{A} B) \\ \Rightarrow & 2P(AB) = 1 – \left[ P(A \bar{B}) + P(\bar{A} B) \right] \\ (3), (4) \Rightarrow & 2P(AB) = 1 – \left\{ \left[ P(A) – P(AB) \right] + \left[ P(B) – P(AB) \right] \right\} \\ \Rightarrow & 2P(AB) = 1 – \left[ P(A) + P(B) – 2 P(AB) \right] \\ \Rightarrow & 2P(AB) = 1 – \left[ P(A) + P(B) \right] + 2 P(AB) \\ \Rightarrow & 0 = 1 – P(A) + P(B) \\ \Rightarrow & P(A) + P(B) = 1 \\ \Rightarrow & \textcolor{springgreen}{\boldsymbol{ A \cup B = \Omega }} \end{aligned}