1723

已知随机事件 ${\displaystyle A,B}$ 均包含于必然事件 ${\displaystyle \mathrm{\Omega }}$, 若 ${\displaystyle 0<\mathbb{P}(A)<1}$, ${\displaystyle 0<\mathbb{P}(B)<1}$, 则

• A. ${\displaystyle \mathbb{P}(B|A)+\mathbb{P}(\bar{B}|A)=\mathbb{P}(A)}$
• B. ${\displaystyle \mathbb{P}(A)+\mathbb{P}(B)\le 1+\mathbb{P}(AB)}$
• C. ${\displaystyle |\mathbb{P}(AB)-\mathbb{P}(A)\mathbb{P}(B)|\le \frac{1}{4}}$
• D. ${\displaystyle \mathbb{P}(B|A)\cdot \mathbb{P}(A|B)=\mathbb{P}(AB)}$

• A 假设 ${\displaystyle A=B}$, 则 ${\displaystyle \mathrm{LHS}=1>\mathbb{P}(A)}$, 所以不正确. 事实上 $$\mathrm{LHS}=\frac{\mathbb{P}(AB)}{\mathbb{P}(A)}+\frac{\mathbb{P}(\bar{B}A)}{\mathbb{P}(A)}=\frac{\mathbb{P}(A)}{\mathbb{P}(A)}=1.$$
• B 由容斥原理 $$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(AB)\le 1,$$ 所以正确.
• C 不妨设 ${\displaystyle \mathbb{P}(A)\ge \mathbb{P}(B)}$. 这样一方面 $$\mathbb{P}(AB)-\mathbb{P}(A)\mathbb{P}(B)\le \mathbb{P}(B)-{\mathbb{P}}^2(B)\le \frac{1}{4},$$ 取等条件为 ${\displaystyle A=B}$, 且 ${\displaystyle \mathbb{P}(A)=\mathbb{P}(B)=\frac{1}{2}}$. 另一方面依据 ${\displaystyle B=(AB)\cup (\bar{A}B)}$, 有$$\begin{array}{rl}\mathbb{P}(AB)-\mathbb{P}(A)\mathbb{P}(B)& =\mathbb{P}(AB)-\mathbb{P}(A)[\mathbb{P}(AB)+\mathbb{P}(\bar{A}B)]\\ & =\mathbb{P}(AB)[1-\mathbb{P}(A)]-\mathbb{P}(A)\mathbb{P}(\bar{A}B)\\ & \ge 0-\mathbb{P}(A)\mathbb{P}(\bar{A})\\ & =-\mathbb{P}(A)[1-\mathbb{P}(A)]\ge -\frac{1}{4}.\end{array}$$ 取等条件为 ${\displaystyle A\cap B=\varnothing }$, 且 ${\displaystyle \mathbb{P}(A)=\mathbb{P}(B)=\frac{1}{2}}$.^[1]
• D 取 ${\displaystyle A=B}$, 则 ${\displaystyle \mathrm{LHS}=1>\mathbb{P}(A)=\mathbb{P}(AB)}$, 所以不正确. 事实上 $$\mathrm{LHS}=\frac{\mathbb{P}(AB)}{\mathbb{P}(A)}\cdot \frac{\mathbb{P}(AB)}{\mathbb{P}(B)}\ne \mathbb{P}(AB).$$

综上, 答案为 BC.