(16)(本小题满分 12 分)
某旅游爱好者计划从 3 个亚洲国家 $A_{1}, A_{2}, A_{3}$ 和 3 个欧洲国家 $B_{1}, B_{2}, B_{3}$ 中选择 2 个国家去旅游。
(I)若从这 6 个国家中任选 2 个,求这 2 个国家都是亚洲国家的概率;
(II)若从亚洲国家和欧洲国家中个任选 1 个,求这 2 个国家包括 $A_{1}$ 但不包括 $B$ 的概率。
(16)(本小题满分 12 分) 某旅游爱好者计划从 3…——2017 高考数学第 16 题答案解析
2017_退役省自主命题 (2017·文)
完整解析 · 逐步详解
【解答】
解:
①由题意知,从 6 个国家中任选两个国家,其一切可能的结果组成的基本事件有:
$\left\{A_{1}, A_{2}\right\},\left\{A_{1}, A_{3}\right\},\left\{A_{2}, A_{3}\right\},\left\{A_{1}, B_{1}\right\},\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\},\left\{A_{2}, B_{1}\right\},\left\{A_{2}, B_{2}\right\},\left\{A_{2}, B_{3}\right\}$, $\left\{A_{3}, B_{1}\right\},\left\{A_{3}, B_{2}\right\},\left\{A_{3}, B_{3}\right\},\left\{B_{1}, B_{2}\right\},\left\{B_{1}, B_{3}\right\},\left\{B_{2}, B_{3}\right\}$ ,共15个
所选两个国家都是亚洲国家的事件所包含的基本事件有:
$\left\{A_{1}, A_{2}\right\},\left\{A_{1}, A_{3}\right\},\left\{A_{2}, A_{3}\right\}$ ,共3个,
则所求事件的概率为:$P=\frac{3}{15}=\frac{1}{5}$
解法二:$P=\frac{C_{3}^{2}}{C_{6}^{2}}=\frac{3}{15}=\frac{1}{5}$
②从亚洲国家和欧洲国家中各任选一个,其一切可能的结果组成的基本事件有:
$\left\{A_{1}, B_{1}\right\},\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\},\left\{A_{2}, B_{1}\right\},\left\{A_{2}, B_{2}\right\},\left\{A_{2}, B_{3}\right\},\left\{A_{3}, B_{1}\right\},\left\{A_{3}, B_{2}\right\},\left\{A_{3}, B_{3}\right\}$,
共 9 个
包括 $A_{1}$ 但不包括 $B_{1}$ 的事件所包含的基本事件有:
$\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\}$ ,共2个,
则所求事件的概率为 $P=\frac{2}{9}$
解法二:$P=\frac{C_{1}^{1} C_{2}^{1}}{C_{3}^{1} C_{3}^{1}}=\frac{2}{9}$
【解答】
解:(I)由题意知,从 6 个国家中任泩寝个国家,其一切可能的结果组成的基本事件有:
$$ \begin{aligned} & \left\{A_{1}, A_{2}\right\},\left\{A_{1}, A_{3}\right\},\left\{A_{2}, A_{3}\right\},\left\{A_{1}, B_{1}\right\},\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\},\left\{A_{2}, B_{1}\right\},\left\{A_{2}, B_{2}\right\},\left\{A_{2}, B_{3}\right\}, \\ & \left\{A_{3}, B_{1}\right\},\left\{A_{3}, B_{2}\right\},\left\{A_{3}, B_{3}\right\},\left\{B_{1}, B_{2}\right\},\left\{B_{1}, B_{3}\right\},\left\{B_{2}, B_{3}\right\} \text {, 共 } 15 \text { 个. } \end{aligned} $$
所选两个国家都是亚洲国家的事件所包含的基本事件有:
$\left\{A_{1}, A_{2}\right\},\left\{A_{1}, A_{3}\right\},\left\{A_{2}, A_{3}\right\}$ ,共3个,
则所求事件的概率为:$P=\frac{3}{15}=\frac{1}{5}$ .
(II)从亚洲国家和欧洲国家中各任选一个,其一切可能的结果组成的基本事件有:
$$ \left\{A_{1}, B_{1}\right\},\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\},\left\{A_{2}, B_{1}\right\},\left\{A_{2}, B_{2}\right\},\left\{A_{2}, B_{3}\right\},\left\{A_{3}, B_{1}\right\},\left\{A_{3}, B_{2}\right\},\left\{A_{3}, B_{3}\right\}, $$
共 9 个。
包括 $A_{1}$ 但不包括 $B_{1}$ 的事件所包含的基本事件有:
$\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\}$ ,共 2 个,
则所求事件的概率为:$P=\frac{2}{9}$ .
【解答】
(本小题满分 12 分)
某旅游爱好者计划从 3 个亚洲国家 $A_{1}, A_{2}, A_{3}$ 和 3 个欧洲国家 $B_{1}, B_{2}, B_{3}$ 中选择 2 个国家去旅游。
(I)若从这 6 个国家中任选 2 个,求这 2 个国家都是亚洲国家的概率;
(II)若从亚洲国家和欧洲国家中各任选 1 个,求这 2 个国家包括 $A_{1}$ 但不包括 $B_{1}$ 的概率.
【解答】
解:(I)由题意知,从 6 个国家中任泩寝个国家,其一切可能的结果组成的基本事件有:
$$ \begin{aligned} & \left\{A_{1}, A_{2}\right\},\left\{A_{1}, A_{3}\right\},\left\{A_{2}, A_{3}\right\},\left\{A_{1}, B_{1}\right\},\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\},\left\{A_{2}, B_{1}\right\},\left\{A_{2}, B_{2}\right\},\left\{A_{2}, B_{3}\right\}, \\ & \left\{A_{3}, B_{1}\right\},\left\{A_{3}, B_{2}\right\},\left\{A_{3}, B_{3}\right\},\left\{B_{1}, B_{2}\right\},\left\{B_{1}, B_{3}\right\},\left\{B_{2}, B_{3}\right\} \text {, 共 } 15 \text { 个. } \end{aligned} $$
所选两个国家都是亚洲国家的事件所包含的基本事件有:
$\left\{A_{1}, A_{2}\right\},\left\{A_{1}, A_{3}\right\},\left\{A_{2}, A_{3}\right\}$ ,共3个,
则所求事件的概率为:$P=\frac{3}{15}=\frac{1}{5}$ .
(II)从亚洲国家和欧洲国家中各任选一个,其一切可能的结果组成的基本事件有:
$$ \left\{A_{1}, B_{1}\right\},\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\},\left\{A_{2}, B_{1}\right\},\left\{A_{2}, B_{2}\right\},\left\{A_{2}, B_{3}\right\},\left\{A_{3}, B_{1}\right\},\left\{A_{3}, B_{2}\right\},\left\{A_{3}, B_{3}\right\}, $$
共 9 个。
包括 $A_{1}$ 但不包括 $B_{1}$ 的事件所包含的基本事件有:
$\left\{A_{1}, B_{2}\right\},\left\{A_{1}, B_{3}\right\}$ ,共 2 个,
则所求事件的概率为:$P=\frac{2}{9}$ .