2020 年研究生入学考试数学一选择题第 7 题解析

题目

A,B,C 为三个随机事件，且 P(A)=P(B)=P(C)=\frac{1}{4},P(AB)=0,P(AC)=P(BC)=\frac{1}{12},A,B,C 中恰有一个发生的概率为 （ ）

A.\frac{3}{4}
B.\frac{2}{3}
C.\frac{1}{2}
D.\frac{5}{12}

方法一

[\frac{1}{4}+\frac{1}{4}-(\frac{1}{12}+\frac{1}{12})]+[\frac{1}{4}-(\frac{1}{12}+\frac{1}{12})]=\frac{5}{12}.

方法二

A\bar{B}\bar{C},B\bar{A}\bar{C},C\bar{B}\bar{A}.

P(A\bar{B} \bar{C})=P(A \bar{B \cup C})=P(A)-P[A(B \cup C)]=P(A)-P(AB+AC)=P(A)-[P(AB)+P(AC)-P(ABC)]=P(A)-P(AB)-P(AC)+P(ABC)=\frac{1}{4}-0-\frac{1}{12}+0=\frac{1}{4}-\frac{1}{12}=\frac{1}{6}.
P(B\bar{A}\bar{C})=P[B(\bar{A \cup C})]=P(B)-P[B(A \cup C)]=P(B)-P(AB+BC)=P(B)-[P(AB)+P(BC)-P(ABC)]=P(B)-P(AB)-P(BC)+P(ABC)=\frac{1}{4}-0-\frac{1}{12}+0=\frac{1}{6}.
P(C\bar{B}\bar{A})=P[C(\bar{B \cup A})]=P(C)-P[C(B \cup A)]=P(C)-P(BC+AC)=P(C)-[P(BC)+P(AC)-P(ABC)]=P(C)-P(BC)-P(AC)+P(ABC)=\frac{1}{4}-\frac{1}{12}-\frac{1}{12}+0=\frac{1}{12}

P(A\bar{B}\bar{C}+B\bar{A}\bar{C}+C\bar{B}\bar{A})=P(A\bar{B}\bar{C})+P(B\bar{A}\bar{C})+P(C\bar{B}\bar{A})=\frac{1}{6}+\frac{1}{6}+\frac{1}{12}=\frac{5}{12}

EOF