设 $A , B$ 为两个随机事件,且 $A$ 与 $B$ 相互独立,已知 $P ( A ) = 2 P ( B ) , P ( A \cup B ) = { \frac { 5 } { 8 } } .$ , 则在事件 $A , B$ 至少有一个发生的条件下, $A , B$ 中恰有一个发生的概率为
答案
$\frac { 4 } { 5 }$
概率论数理统计
📋 解题步骤
1
分析题意,确定思路
▼
$P ( A { \bar { B } } ) + P ( { \overline { { A } } } B ) = P ( A ) - P ( A B ) + P ( B ) - P ( A B )$
$$
= P (A) + P (B) - 2 P (A) P (B)
$$
$$
P (A \cup B) = P (A) + P (B) - P (A B) = \frac {5}{8}, \Rightarrow 3 P (B) - 2 P ^ {2} (B) = \frac {5}{8}
$$
2
建立方程或引用定理
▼
$$
2 4 P (B) - 1 6 P ^ {2} (B) = 5, 1 6 P ^ {2} (B) - 2 4 P (B) + 5 = 0
$$
$$
\Rightarrow (4 P (B) - 1) (4 P (B) - 5) = 0
$$
$$
P (B) = \frac {1}{4}, P (A) = \frac {1}{2}
$$
$$
\begin{array}{l} P (A \bar {B}) + P (\bar {A} B) = P (A) - P (A B) + P (B) - P (A B) = P (A) + P (B) - 2 P (A) P (B) \\ = \frac {1}{2} + \frac {1}{4} - 2 \times \frac {1}{4} \times \frac {1}{2} = \frac {1}{2} \\ \end{array}
$$