《2025年暑假作业甘肃少年儿童出版社高一数学人教版》

11. 甲、乙两队进行足球比赛,若两队战平的概率是$ \frac{1}{4} $,乙队胜的概率是$ \frac{1}{3} $,则甲队胜的概率是$\frac{5}{12}$.

答案： 11.$\frac{5}{12}$

12. 从甲、乙等5名同学中随机选3名参加社区服务工作,则甲、乙都入选的概率为$\frac{3}{10}$.

答案： 12.$\frac{3}{10}$

13. 试验E:箱子里有3双不同的手套,随机拿出2只,记随机事件A为“拿出的手套配不成对”;随机事件B为“拿出的是同一只手上的手套”;随机事件C为“拿出的手套一只是左手的,一只是右手的,但配不成对”.
(1)写出试验E的样本空间Ω,并指出样本点的个数;
(2)分别用样本点表示随机事件A、随机事件B、随机事件C,并指出每个随机事件的样本点的个数;
(3)写出$ A\cap B,B\cap C,A\cap C,B\cup C $.

答案： 13.【解析】
(1)分别设3双手套为$a_{1},a_{2},b_{1},b_{2},c_{1},c_{2}$，其中$a_{1},b_{1},c_{1}$分别代表左手的3只手套，$a_{2},b_{2},c_{2}$分别代表右手的3只手套。试验$E$的样本空间$\Omega = \{ (a_{1},a_{2}),(a_{1},b_{1}),(a_{1},b_{2}),(a_{1},c_{1}),(a_{1},c_{2}),(a_{2},b_{1}),(a_{2},b_{2}),(a_{2},c_{1}),(a_{2},c_{2}),(b_{1},b_{2}),(b_{1},c_{1}),(b_{1},c_{2}),(b_{2},c_{1}),(b_{2},c_{2}),(c_{1},c_{2}) \}$，样本点的个数为15。
(2)随机事件$A = \{ (a_{1},b_{1}),(a_{1},b_{2}),(a_{1},c_{1}),(a_{1},c_{2}),(a_{2},b_{1}),(a_{2},b_{2}),(a_{2},c_{1}),(a_{2},c_{2}),(b_{1},c_{1}),(b_{1},c_{2}),(b_{2},c_{1}),(b_{2},c_{2}) \}$，样本点的个数为12。
随机事件$B = \{ (a_{1},b_{1}),(a_{1},c_{1}),(a_{2},b_{2}),(a_{2},c_{2}),(b_{1},c_{1}),(b_{2},c_{2}) \}$，样本点的个数为6。
随机事件$C = \{ (a_{1},b_{2}),(a_{1},c_{2}),(a_{2},b_{1}),(a_{2},c_{1}),(b_{1},c_{2}),(b_{2},c_{1}) \}$，样本点的个数为6。
(3)$A \cap B = \{ (a_{1},b_{1}),(a_{1},c_{1}),(a_{2},b_{2}),(a_{2},c_{2}),(b_{1},c_{1}),(b_{2},c_{2}) \}$，$B \cap C = \varnothing$，$A \cap C = \{ (a_{1},b_{2}),(a_{1},c_{2}),(a_{2},b_{1}),(a_{2},c_{1}),(b_{1},c_{2}),(b_{2},c_{1}) \}$，$B \cup C = \{ (a_{1},b_{1}),(a_{1},b_{2}),(a_{1},c_{1}),(a_{1},c_{2}),(a_{2},b_{1}),(a_{2},b_{2}),(a_{2},c_{1}),(a_{2},c_{2}),(b_{1},c_{1}),(b_{1},c_{2}),(b_{2},c_{1}),(b_{2},c_{2}) \}$。

14. 有A,B,C,D四位贵宾,应分别坐在a,b,c,d四个席位上,现在这四人均未留意,在四个席位上随便就座时,
(1)求这四人恰好都坐在自己的席位上的概率.
(2)求这四人恰好都没坐在自己的席位上的概率.

答案：
14.【解析】将$A,B,C,D$四位贵宾就座情况用下面图形表示出来：

如图所示，本题中的样本点的总数为24。
(1)设事件$A$为“这四人恰好都坐在自己的席位上”，则事件$A$只包含1个样本点，所以$P(A)=\frac{1}{24}$。
(2)设事件$B$为“这四个人恰好都没有坐在自己席位上”，则事件$B$包含9个样本点，所以$P(B)=\frac{9}{24}=\frac{3}{8}$。