摘要： 在一个不透明的盒子里.装有四个分别标有数字1.2.3.4的小球.它们的形状.大小.质地等完全相同.小明先从盒子里随机取出一个小球.记下数字为x,放回盒子摇匀后.再由小华随机取出一个小球.记下数字为y. (1)用列表法或画树状图表示出(x.y)的所有可能出现的结果, (2)求小明.小华各取一次小球所确定的点(x.y)落在反比例函数的图象上的概率, (3)求小明.小华各取一次小球所确定的数x.y满足的概率. [答案]解:(1) x y 1 2 3 4 1 (1.1) (2.1) (3.1) (4.1) 2 (1.2) (2.2) (3.2) (4.2) 3 (1.3) (2.3) (3.3) (4.3) 4 (1.4) (2.4) (3.4) (4.4) ······························································································································································ 3分 (2)可能出现的结果共有16个.它们出现的可能性相等.·········································· 4分 满足点(x.y)落在反比例函数的图象上(记为事件A)的结果有3个.即. (4.1). 所以P(A)=.·································································································································· 7分 (3)能使x.y满足(记为事件B)的结果有5个.即..所以P(B)= 9分

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